1.अवकल समीकरण का विशिष्ट समाकल (Particular Integral of DE),पूरक फलन और विशिष्ट समाकल (Complementary Function and Particular Integral):

अवकल समीकरण का विशिष्ट समाकल (Particular Integral of DE) की यह विधि प्रयोग में लाना कठिन है।अतः इस अवस्था में पिछली विधियों से भी विशिष्ट समाकल ज्ञात किया जा सकता है।
\frac{1}{f(D)}(x,v) का मान ज्ञात करना जहाँ V, x का कोई फलन है (Evaluate \frac{1}{f(D)}(x,v), Where V is a function of x.)
उत्तरोत्तर अवकलन (Successive Differentiation) से हम देखते हैं किः

D(x V)=x D V+V \\ D^{2}(x V)=x D^{2} V+2 D V \\ D^{3}(x V)=x D^{3} V+3 D^{2} V \\ \cdots \cdots \cdots \cdots \\ D^{n}(x V)=x D^{n} V+x D^{n-1} V=x D^{n} V+ \frac{d}{d D} D^{n} V
अतः व्यापक रूप से:

f(D) (x V)=x f(D) V+\left\{\frac{d}{d D} f(D) \right\} V \cdots(1)
इससे हम यह अर्थ निकाल सकते है कि

\frac{1}{f(D)}(x V)=x f(D) V +\left \{ \frac{d}{d D} \frac{1}{f(D)} \right \} V \cdots(2)
जहाँ \frac{d}{d D} \cdot \frac{1}{f(D)}=-f(D) \frac{1}{\left[f(D)\right]^{2}}
यह देखने के लिए कि उपर्युक्त परिणाम सत्य है। हम इस पर f(D) की संक्रिया करते हैं तो दाहिने पक्ष का मान होगा:

f(D)\left[x \frac{1}{f(D)} V+\left\{\frac{d}{d D} \cdot \frac{1}{f(D)} V\right\}\right]=x f(D) \frac{1}{f(D)} V+\left\{\frac{d}{d D} f(D)\right\} \frac{1}{f(D)} V+f(D)\left[\frac{d}{dD} \cdot \frac{1}{f(D)}\right]V

[(1) से]
=x V+f^{\prime}(D) \frac{1}{f(D)} V-f(D) f^{\prime}(D) \frac{1}{[f(D)]^{2}} V[क्रमविनिमेय के नियम से]

=x V+f^{\prime}(D) \frac{1}{f(D)} V-f^{\prime}(D) \frac{1}{f(D)} V=x V
यह वही है जो (2) के वाम पक्ष पर f(D) से संक्रिया करने पर प्राप्त होता है।अतः परिणाम (2) सत्य है।
टिप्पणी:यह बात ध्यान देने योग्य है कि सूत्र (1) हमको लैबनीज-प्रमेय के प्रयोग से ज्ञात हुआ है जो कि उत्तरोत्तर अवकलन का व्यापक रूप है।अतः हम देखते हैं कि

D^{n} \left(x^{2} V\right)=x^{2} D^{n} V+2 x n D^{n-1} V+2 n \frac{(n-1)}{1.2} D^{n-2} V \\ =x^{2} D^{n} V+2 x\left(\frac{d}{d D} D^{n}\right) V+\left(\frac{d^{2}}{d D^{2}} D^{n}\right) V
व्यापक रूप में:

f(D)\left(x^{2} V\right)=x^{2}f(D) V+2 x\left\{\frac{d}{d D} f(D)\right\} V+\left\{\frac{d^{2}}{d D^{2}} f(D)\right\} V
अतः \frac{1}{f(D)} x^{2} V=x^{2} \frac{1}{f(D)} V+2 x\left\{\frac{d}{d D} \cdot \frac{1}{f(D)}\right\} V+\left\{\frac{d^{2}}{d D^{2}} \cdot \frac{1}{f(D)}\right\} V
इस प्रकार हम के लिए सूत्र ज्ञात कर सकते हैं परन्तु वह प्रयोग में लाना कठिन सिद्ध होगा।ऐसी अवस्था में v e^{ax} का पिछली विधि से सरल करना उपयुक्त होगा।
आपको यह जानकारी रोचक व ज्ञानवर्धक लगे तो अपने मित्रों के साथ इस गणित के आर्टिकल को शेयर करें।यदि आप इस वेबसाइट पर पहली बार आए हैं तो वेबसाइट को फॉलो करें और ईमेल सब्सक्रिप्शन को भी फॉलो करें।जिससे नए आर्टिकल का नोटिफिकेशन आपको मिल सके । यदि आर्टिकल पसन्द आए तो अपने मित्रों के साथ शेयर और लाईक करें जिससे वे भी लाभ उठाए । आपकी कोई समस्या हो या कोई सुझाव देना चाहते हैं तो कमेंट करके बताएं।इस आर्टिकल को पूरा पढ़ें।

Also Read This Article:-CF and PI of Differential Equation

2.अवकल समीकरण का विशिष्ट समाकल के साधित उदाहरण (Particular Integral of DE Solved Examples):

Example:1. \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x \sin x
Solution: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}-2 m+1=0 \\ \Rightarrow \left(m-1\right)^{2}=0\\ \Rightarrow m=1,1

C.F.=(C_{1}+C_{2} x) e^{x}

P.I=\frac{1}{D^{2}-2 D+1} x \sin x \\ =x \frac{1}{\left(D^{2}-2 D+1\right)} \sin x+\left[\frac{d}{dD}\frac{1}{ D^{2}-2 D+1}\right] \sin x \\ =x \cdot \frac{1}{i^{2}-2 D+1} \sin x+\left[\frac{d}{d D} \cdot \frac{1}{(D-1)^{2}}\right] \sin x \\ =x\left(-\frac{1}{2 D}\right) \sin x+\left[-\frac{2}{(D-1)^{3}}\right] \sin x \\ =\frac{1}{2} x \cos x-2\left[\frac{1}{D^{3}-3 D^{2}+3 D-1}\right] \sin x \\ =\frac{1}{2} x \cos x-2\left[\frac{1}{D i^{2}-3 i^{2}+3 D-1}\right] \sin x \\ =\frac{1}{2} x \cos x-2\left[\frac{1}{-D+3+3 D-1}\right] \sin x \\ =\frac{1}{2} x \cos x-2\left[\frac{1}{2 D+2}\right] \sin x \\ =\frac{1}{2} x \cos x-\left[\frac{1}{D+1}\right] \sin x \\ =\frac{1}{2} x \cos x-\left[\frac{D-1}{D^{2}-1}\right] \sin x \\ =\frac{1}{2} x \cos x-\left[\frac{(D-1) \sin x}{i^{2}-1}\right] \\ =\frac{1}{2} x \cos x+\frac{1}{2}(\cos x-\sin x) \\ P.I.=\frac{1}{2}(x \cos x+\cos x-\sin x)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=\left(C_{1}+C_{2} x\right) e^{x}+\frac{1}{2}(x \cos x+\cos x-\sin x)
Example:2. \left(D^{2}+4\right) y=x \sin x
Solution: \left(D^{2}+4\right) y=x \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+4=0 \Rightarrow m=\pm 2 i

C.F.=C_{1} \cos 2 x+C_{2} \sin 2 x

P.I.=\frac{1}{\left(D^{2}+4\right)}(x \sin x) \\ =x \frac{1}{D^{2}+4} \sin x+\left[\frac{d}{d D} \cdot \frac{1}{D^{2}+4} \right] \sin x \\ =\frac{x \sin x}{i^{2}+4}+\left[\frac{-2 D}{\left(D^{2}+4\right)^{2}}\right] \sin x \\ =\frac{x \sin x}{-1+4}+\left[\frac{-2 D(\sin x)}{\left(i^{2}+4\right)^{2}}\right] \\ =\frac{1}{3} x \sin x-2\left[\frac{\cos x}{(-1+4)^{2}}\right] \\ \Rightarrow P.I.=\frac{1}{3} x \sin x-\frac{2}{9} \cos x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} \cos 2 x+C_{2} \sin 2 x+\frac{1}{3} x \sin x-\frac{2}{9} \cos x
Example:3. \frac{d^{2} y}{d x^{2}}+4 y=x \cos x
Solution: \frac{d^{2} y}{d x^{2}}+4 y=x \cos x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+4=0 \Rightarrow m=\pm 2 i

C.F.=C_{1} \cos 2 x+C_{2} \sin 2 x

P.I.=\frac{1}{D^{2}+4} x \cos x \\ =x \frac{1}{D^{2}+4} \cos x+\left(\frac{d}{d D} \cdot \frac{1}{D^{2}+4}\right) \cos x \\ =\frac{x \cos x}{i^{2}+4}+\left[\frac{-2 D}{\left(D^{2}+4\right)^{2}}\right] \cos x \\ =\frac{1}{3} x \cos x-2\left[\frac{D(\cos x)}{\left(i^{2}+4\right)^{2}}\right] \\ =\frac{1}{3} x \cos x+\frac{2 \sin x}{(-1+4)^{2}} \\ \Rightarrow P.I.=\frac{1}{3} x \cos x+\frac{2}{9} \sin x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} \cos 2 x+C_{2} \sin 2 x+\frac{1}{3} x \cos x+\frac{2}{9} \sin x
Example:4. \left(D^{2}-1\right) y=x \sin x
Solution: \left(D^{2}-1\right) y=x \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{4}-1=0 \Rightarrow(m-1)(m+1)\left(m^{2}+1\right)=0 \\ \Rightarrow m=1,-1, \pm i

C.F.=C_{1} e^{x}+C_{2} e^{-x}+C_{3} \cos x +C_{4} \sin x

P.I=\frac{1}{\left(D^{4}-1\right)} x \sin x \\ = \text{I.P.} \frac{1}{\left(D^{4}-1\right)} x e^{i x} \\ =\text{I.P. } e^{i x} \frac{1}{(D+i)^{4}-1} x \\ =\text{I.P. } e^{i x} \frac{1}{D^{4}+4 D^{3} i-6 D^{2}-4 D i+1-1} x \\ =\text{I.P. } e^{i x} \frac{1}{D^{4}+4 D^{3} i-6 D^{2}-4 D i} x \\ =\text{I.P. } e^{i x} \cdot \frac{1}{-4 D i} \cdot \frac{1}{\left[1-\left(\frac{D^{3}}{4 i}+D^{2}-\frac{3 D}{2i}\right)\right]} \\ =\text{I.P.} e^{i x} \left(\frac{1}{-4 D i} \right)\left[1-\left(\frac{D^{3}}{4 i}+D^{2}-\frac{3 D}{2 i}\right)\right]^{-1} x \\ =\text{I.P.} e^{i x}\left(\frac{1}{-4 D i}\right)\left[1+\frac{D^{3}}{4 i}+D^{2}-\frac{3D}{2 i}+\left(\frac{D^{3}}{4 i}+D^{2}- \frac{3D}{2 i}\right)+\cdots \right] x \\ =\text{I.P. } e^{ix}\left(\frac{1}{-4 D i}\right) \left[1+D^{2} -\frac{3 D}{2 i}-\frac{9 D^{2}}{4}+\cdots \right]x\\ =\text{I.P. } \frac{e^{i x}}{(-4 i)}\left[\frac{1}{D}-\frac{3}{2 i}-\frac{5 D}{4}+\cdots \right]x\\ =\text{I.P. } \frac{(\cos x+i \sin x)}{4}(i)\left[\frac{x^{2}}{2}-\frac{3 x}{2 i}-\frac{5}{4}\right] \\ =\text{I.P. } \frac{(\cos x+ i \sin x )}{4}\left ( \frac{i x^{2}}{2}-\frac{3x}{2}-\frac{5i}{4} \right ) \\ =\frac{1}{4} \text{I.P. }\left[-\frac{3}{2} x \cos x-\frac{x^{2}}{2} \sin x+\frac{5}{4} \sin x+i \left(\frac{x^{2}}{2} \cos x-\frac{5}{4} \cos x-\frac{3}{2} x \sin x\right)\right] \\ =\frac{1}{4}\left(\frac{x^{2}}{2} \cos x-\frac{5}{4} \cos x-\frac{3}{2} x \sin x\right) \\ \Rightarrow \text{P. I.}=\frac{x^{2}}{8} \cos x-\frac{3}{8} x \sin x-\frac{5}{8} \cos x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y =C_{1} e^{x}+C_{2} e^{-x}+C_{3} \cos x+C_{4} \sin x +\frac{1}{8}\left(x^{2} \cos x-3 x \sin x\right)
Example:5. \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=x \cos x
Solution: \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=x \cos x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+2 m+1=0 \\ \Rightarrow(m+1)^{2}=0 \Rightarrow m=-1,-1

C.F.=\left(C_{1}+C_{2} x\right) e^{-x}

P.I=\frac{1}{D^{2}+2 D+1} x \cos x \\ =x \frac{1}{D^{2}+2 D+1} \cdot \cos x+\left[\frac{d}{d D} \frac{1}{(D+1)^{2}}\right] \cos x \\ =x \cdot \frac{1}{i^{2}+2 D+1} \cos x-\left[\frac{2}{(D+1)^{3}}\right] \cos x \\ =\frac{x}{2}\left(\frac{1}{D}\right) \cos x-2\left[\frac{1}{D^{3}+3 D^{2}+3 D+1}\right] \cos x \\ =\frac{x}{2} \sin x-2\left[\frac{1}{D i^{2}+3 i^{2}+3 D+1}\right] \cos x\\ =\frac{x}{2} \sin x-2\left[\frac{1}{-D-3+3 D+1}\right] \cos x \\ =\frac{x}{2} \sin x-2\left[\frac{1}{2 D-2}\right] \cos x \\ =\frac{x}{2} \sin x-\left[ \frac{D+1}{D^{2}-1}\right] \cos x \\ =\frac{x}{2} \sin x-\frac{(D+1) \cos x}{i^{2}-1} \\ =\frac{x}{2} \sin x-\frac{(-\sin x+\cos x)}{-2} \\ =\frac{x}{2} \sin x-\frac{1}{2} \sin x+\frac{1}{2} \cos x \\ \Rightarrow \text{P. I.}=\frac{x}{2} \sin x-\frac{1}{2} \sin x+\frac{1}{2} \cos x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y =\left(C_{1}+C_{2} x\right) e^{-x}+\frac{1}{2}(x-1) \sin x+\frac{1}{2} \cos x
Example:6. \frac{d^{3}y}{d x^{3}}+y=x \cos 2 x
Solution: \frac{d^{3}y}{d x^{3}}+y=x \cos 2 x
इसका सहायक समीकरण (A.E.) होगा:

m^{3}+1=0 \Rightarrow(m+1)\left(m^{2}-m+1\right)=0 \\ \Rightarrow m=-1, m^{2}-m+1=0 \\ \Rightarrow m=\frac{1 \pm \sqrt{(-1)^{2}-4 \times 1 \times 1}}{2} \\ \Rightarrow m=\frac{1 \pm \sqrt{1-4}}{2} \\ \Rightarrow m=\frac{1 \pm \sqrt{3} i}{2} \\ m=-1,\left(\frac{1 \pm 3 i}{2}\right)

C.F=C_{1} e^{x}+e^{\frac{x}{2}}\left[C_{2} \cos \left(\frac{\sqrt{3}}{2} x\right)+C_{3} \sin \left(\frac{\sqrt{3} x}{2}\right)\right]

P.I.=\frac{1}{D^{3}+1} x \cos 2 x \\ =x \frac{1}{D^{3}+1} \cos 2 x+\left[\frac{d}{d D} \cdot \frac{1}{\left(D^{3}+1\right)}\right] \cos 2 x\\ =x \frac{1}{D(2 i)^{2}+1} \cos 2 x+\frac{-3 D^{2}}{\left(D^{3}+1\right)^{2}} \cos 2 x\\ =x \frac{1}{1-4 D} \cos 2 x-\frac{3 D^{2}}{(2i)^{6}+2D (2i)^{2}+1} \cos 2 x\\ =x \frac{(1+4 D)}{1-16 D^{2}} \cos 2 x-\frac{3 D^{2}}{64 i^{6}+8Di^{2}+1} \cos 2 x\\ =\frac{x(1+4 D) \cos 2 x}{1-16(2 i)^{2}}-\frac{3 D^{2}}{64 i^{6}+8 D i^{2}+1} \cos 2 x\\ =\frac{x(1+4 D) \cos 2 x}{1-16 \times 4i^{2}}-\frac{3 D^{2}}{-64 -8 D+1} \cos 2 x\\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)-\frac{3 D^{2}}{(-63-8 D)} \cos 2 x \\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)+\frac{3 D^{2}(8 D-63)}{64 D^{2}-63^{2}} \cos 2 x \\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)+\frac{3\left(8 D^{3}-63 D^{2}\right) \cos 2 x}{64(2 i)^{2}-3969}\\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)+\frac{3\left(64 \sin 2x-252 \cos 2x\right) \cos 2 x}{64 \times 4i^{2}-3969}\\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)+\frac{(192 \sin 2 x+756 \cos 2x)}{-256-3969}\\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)-\frac{(192 \sin 2 x+756 \cos 2 x)}{4225}\\ =\frac{x}{65}(\cos 2 x-8 \sin 2 x)-\frac{1}{(65)^{2}}(192 \sin 2 x+756 \cos 2 x)\\ =\frac{1}{(65)^{2}}[(65 x-756) \cos 2 x-(192+520 x) \sin 2 x] \\ \Rightarrow \text{P. I.}=\frac{1}{(65)^{2}}[(65 x-756) \cos 2 x-(192+520 x) \sin 2 x]
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} e^{-x}+e^{\frac{x}{2}}\left[C_{2} \cos \left(\frac{\sqrt{3}}{2} x\right)+C_{3} \sin \left(\frac{\sqrt{3}}{2} x\right)\right]+\frac{1}{(65)^{2}}[(65 x-756) \cos 2 x-(192+520 x) \sin 2 x]
Example:7. \left(D^{2}+4\right) y=x \sin 2 x
Solution: \left(D^{2}+4\right) y=x \sin 2 x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+4=0 \Rightarrow m=\pm 2 i

C.F.=C_{1} \cos 2 x+C_{2} \sin 2 x

P.I.=\frac{1}{\left(D^{2}+4\right)}x \sin 2 x=x \frac{1}{D^{2}+4} \sin 2 x+ \left[\frac{d}{dD} \cdot \frac{1}{D^{2}+4} \right] \sin 2 x \\ =x\left(-\frac{x}{4} \cos 2 x\right)-\frac{2 D}{\left(D^{2}+4\right)^{2}} \sin 2 x \\ =\frac{-x^{2}}{4} \cos 2 x-\frac{4 \cos 2 x}{\left(D^{2}+4\right)^{2}} \\ =\frac{-x^{2}}{4} \cos 2 x -\frac{4}{\left(D^{2}+4\right)} \cdot \left(\frac{1}{D^{2}+4} \cos 2 x\right)\\ =-\frac{x^{2}}{4} \cos 2 x-\frac{4}{D^{2}+4} \cdot \left(\frac{x}{4} \sin 2 x\right) \\ \frac{1}{D^{2}+4}(x \sin 2 x)=-\frac{1}{4} x^{2} \cos 2 x - \frac{1}{\left(D^{2}+4\right)}(x \sin 2 x)\\ \frac{2}{D^{2}+4}(x \sin 2 x)=-\frac{1}{4} x^{2} \cos 2 x \\ \Rightarrow \text{P.I.}=-\frac{1}{8} x^{2} \cos 2 x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

y=C_{1} \cos 2 x+C_{2} \sin 2 x-\frac{1}{8} x^{2} \cos 2 x
वैकल्पिक रीति (Alternate Method):

P.I.=\frac{1}{\left(D^{2}+4\right)} x \sin 2x\\ =\text{I.P.} \frac{1}{\left(D^{2}+4\right)} x e^{2 i x}\\ =\text{I.P. } e^{2 i x} \frac{1}{\left(D+2i\right)^{2}+4} x\\ =\text{I.P. } e^{2 i x} \frac{x}{D^{2}+4 D i+4 i^{2}+4} x\\ =\text{I.P.} e^{2 i x} \frac{1}{D^{2}+4 D i} x\\ =\text{I.P. } e^{2 i x} \frac{1}{4 D i\left[1+\frac{D}{4 i}\right]} x\\ =\text{I.P. } e^{2 i x} \frac{1}{4 D i}\left[1+\frac{D}{4 i}\right]^{-1} x \\ =\text{I.P. } e^{2 i x} \frac{(-i)}{4 D i} \left[1-\frac{D}{4 i}-\frac{D^{2}}{16}+ \cdots \right] x\\ =\text{I.P. } \frac{e^{2 i x}}{4}(-i)\left[\frac{1}{D}-\frac{1}{4 i}-\frac{D}{16}+\cdots \right]x\\ =\frac{1}{4} \text{I.P. } e^{2i x}\left[\frac{-i}{D}+\frac{1}{4}+\frac{D i}{16}+\cdots\right] x\\ =\frac{1}{4} \text{I.P. } (\cos 2 x+i \sin 2 x)\left(-\frac{x^{2} i}{2}+\frac{x}{4}+\frac{i}{16}\right) \\ =\frac{1}{4} \text{I.P. } [\left(\frac{x}{4} \cos 2 x+\frac{x^{2}}{2} \sin 2 x-\frac{1}{16} \sin 2 x\right)+i\left(-\frac{x^{2}}{2} \cos 2 x+\frac{1}{16} \cos 2 x+\frac{1}{4} x \sin 2 x\right)] \\ =\frac{1}{4}\left(-\frac{x^{2}}{2} \cos 2 x+\frac{1}{16} \cos 2 x+\frac{1}{4} x \sin 2 x\right) \\ \Rightarrow \text{P.I.}=-\frac{1}{8} x^{2} \cos 2 x+\frac{1}{16} x \sin 2x+\frac{1}{64} \cos 2 x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} \cos 2 x+C_{2} \sin 2 x-\frac{1}{8} x^{2} \cos 2 x+\frac{1}{16} x \sin 2x
Example:8. \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x e^{x} \sin x
Solution: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x e^{x} \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}-2 m+1=0 \Rightarrow(m-1)^{2}=0 \\ \Rightarrow m=1,1

C.F.=\left(C_{1}+C_{2}x\right) e^{x}

P.I.=\frac{1}{D^{2}-2 D+1} x e^{x} \sin x \\ = \frac{1}{(D-1)^{2}} x e^{x} \sin x \\ = e^{x} \frac{1}{(D+1-1)^{2}} x \sin x \\= e^{x} \cdot \frac{1}{D^{2}} x \sin x \\ = e^{x}\left[x \frac{1}{D^{2}} \sin x+\left[\frac{d}{d D} \frac{1}{D^{2}}\right] \sin x\right] \\ = e^{x}\left[-x \sin x-\frac{2}{D^{3}} \sin x\right] \\ =e^{x}[-x \sin x-2 \cos x] \\ \Rightarrow \text{P.I.}=-e^{x}(x \sin x+2 \cos x)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=\left(C_{1}+C_{2}x\right) e^{x}-e^{x}(x \sin x+2 \cos x)
Example:9. \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x e^{x} \cos x
Solution: \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y=x e^{x} \cos x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}-2 m+1=0 \Rightarrow(m-1)^{2}=0 \\ \Rightarrow m=1,1

C.F.=\left(C_{1}+C_{2} x\right) e^{x}

P.I.=\frac{1}{D^{2}-2 D+1} x e^{x} \cos x \\ =\frac{1}{(D-1)^{2}} x e^{x} \cos 2 x \\ =e^{x} \frac{1}{(D+1-1)^{2}} x \cos 2 x \\ =e^{x} \cdot \frac{1}{D^{2}} x \cos 2 x \\ =e^{x}\left[x \cdot \frac{1}{D^{2}} \cos 2 x+\left[\frac{d}{d D} \cdot \frac{1}{D^{2}}\right] \cos 2 x\right] \\ =e^{x}\left[x\left(-\frac{1}{4} \cos 2 x\right)-\frac{2}{D^{3}} \cos 2 x\right] \\ =e^{x}\left[-\frac{x}{4} \cos 2 x+\frac{1}{4} \sin 2 x\right] \\ \Rightarrow \text{P.I.} =\frac{1}{4} e^{x}(\sin 2 x-x \cos 2 x)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=\left(C_{1}+C_{2} x\right) e^{x}+\frac{1}{4} e^{x}(\sin 2 x-x \cos 2 x)
Example:10. \left(D^{2}+y\right) y=x \sin ^{2} x
Solution: \left(D^{2}+y\right) y=x \sin ^{2} x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+4=0 \Rightarrow m=\pm 2 i

C.F.=C_{1} \cos 2 x+C_{2} \sin 2 x

P.I.=\frac{1}{D^{2}+4}\left(x \sin ^{2} x\right) \\ =\frac{1}{2} \cdot \frac{1}{\left(D^{2}+4\right)} x(1-\cos 2 x) \\ =\frac{1}{2} \cdot \frac{1}{\left(D^{2}+4\right)} x-\frac{1}{2} \cdot \frac{1}{D^{2}+4}(x \cos 2 x) \\ =\frac{1}{8}\left(1+\frac{D^{2}}{4}\right)^{-1} x- \text{R.P. }\left(\frac{1}{2}\right) \frac{1}{D^{2}+4} e^{2 i x} x \\ =\frac{1}{8}\left[1-\frac{1}{4} D^{2}+\cdot \cdot\right]-\frac{1}{2} \text{R.P. } e^{2 i x} \frac{1}{(D+2 i)^{2}+4} x \\ =\frac{1}{8} x-\frac{1}{2} \text{R.P. } e^{2 i x} \cdot \frac{1}{D^{2}+4 D i+4 i^{2}+4} x\\ =\frac{1}{8} x-\frac{1}{2} \text{R.P. } e^{2 i x} \cdot \frac{1}{\left(D^{2}+4 D i\right)} x \\ =\frac{1}{8} x-\frac{1}{2} \text{R.P. } \frac{1}{4 D i\left(1+\frac{D}{4 i}\right)} x \\ =\frac{1}{8} x-\frac{1}{2} \text { R.P. of } e^{2 i x}\left(-\frac{i}{4 D}\right)\left(1+\frac{D}{4 i}\right)^{-1} x \\ =\frac{1}{8} x-\frac{1}{8} \text { R.P. of } e^{2 i x}\left(-\frac{i}{D}\right)\left(1-\frac{D}{4 i}-\frac{D^{2}}{16}+\cdots \right)x \\ =\frac{1}{8} x-\frac{1}{8} \text { R.P. of } e^{2 ix}(-i)\left(\frac{1}{D}-\frac{1}{4 i}-\frac{D}{16}+ \cdots \right) x \\ =\frac{1}{8} x-\frac{1}{8} \text { R.P. of } (\cos 2 x+i \sin 2 x)\left(-\frac{i x^{2}}{2}-\frac{1}{4} x-\frac{i}{16}\right) \\ =\frac{1}{8} x-\frac{1}{8} \text { R.P. of } \left(-\frac{1}{4} x \cos 2 x+\frac{x^{2}}{2} \sin 2 x+\frac{1}{16} \sin 2 x\right) +i\left(-\frac{x^{2}}{2} \cos 2 x-\frac{1}{16} \cos 2 x-\frac{x}{4} \sin 2 x \right) \\ =\frac{1}{8} x-\frac{1}{8} \left ( -\frac{1}{4} x \cos 2 x+\frac{x^{2}}{2} \sin 2 x+\frac{1}{16} \sin 2 x\right ) \\ =\frac{1}{8} x-\frac{1}{32} x \cos 2 x-\frac{1}{16} x^{2} \sin 2 x-\frac{1}{128} \sin 2 x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} \cos 2 x+C_{2} \sin 2 x+\frac{1}{8} x-\left(\frac{1}{16}\right) x^{2} \sin 2 x-\left(\frac{1}{32}\right) x \cos 2 x
Example:11. \left(D^{4}+2 D^{2}+1\right) y=x^{2} \cos x
Solution: \left(D^{4}+2 D^{2}+1\right) y=x^{2} \cos x
इसका सहायक समीकरण (A.E.) होगा:

m^{4}+2 m^{2}+1=0 \Rightarrow\left(m^{2}+1\right)^{2}=0\\ \Rightarrow m=\pm i, \pm i

C.F.=\left(C_{1}+C_{2} x\right) \cos x+\left(C_{3}+C_{4} x\right) \sin x

P.I. =\frac{1}{\left(D^{2}+1\right)^{2}} x^{2} \cos x\\ =\text{R.P. of } \frac{1}{\left(D^{2}+1\right)^{2}} x^{2} e^{i x}\\ =\text{R.P. of } e^{i x} \frac{1}{\left[(D+i)^{2}+1\right]^{2}} x^{2}\\ =\text{R.P. of } e^{i x} \frac{1}{\left(D^{2}+2 D i+i^{2}+1\right)^{2}} x^{2}\\=\text{R.P. of } e^{i x} \frac{1}{\left(D^{2}+2 D i\right)^{2}} x^{2} \\ =\text{R.P. of } e^{i x}\left(\frac{1}{4 D^{2} i^{2}}\right) \frac{1}{\left(1+\frac{D}{2 i}\right)^{2}} x^{2} \\=\text{R.P. of } \frac{e^{i x}}{-4 D}\left(1+\frac{D}{2 i}\right)^{-2} x^{2} \\ =-\frac{1}{4} \text{R.P. of } e^{i x}\left(\frac{1}{D^{2}}\right)\left[1-\frac{2 D}{2 i}-\frac{3 D^{2}}{4}+\frac{4 D^{2}}{8 i}+\cdots \right]x^{2} \\ =-\frac{1}{4} \text{R.P. of } e^{i x}\left[\frac{1}{D^{2}}+\frac{i}{D}-\frac{3}{4}-\frac{1}{2} D i+\cdots\right] x^{2} \\ =-\frac{1}{4} \text{R.P. of } e^{i x}\left[\frac{x^{4}}{12}+\frac{x^{3}}{3} i-\frac{3}{4} x^{2}-x \right] \\ =-\frac{1}{4} \text{R.P. of } (\cos x+i \sin x)\left(\frac{x^{4}}{12}+\frac{x^{3}}{3} i-\frac{3}{4} x^{2}-x i\right)\\ =-\frac{1}{4} \text{R.P. of } [( \frac { x ^ { 4 } } { 1 2 } \cos x - \frac{3}{4} x^{2} \cos x - \frac {x^{3}}{3} \sin x + x \sin x )+i(\frac{x^{3}}{3} \cos x-x \cos x+\frac{x^{4}}{12} \sin x-\frac{3}{4} x^{2} \sin x)]\\ \text{P.I.}=-\frac{1}{48} \left(x^{4}-9 x^{2}\right) \cos x+\frac{1}{12} x^{3} \sin x-\frac{1}{4} x \sin x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=\left(C_{1}+C_{2} x\right) \cos x+\left(C_{3}+C_{4} x \right) \sin x -\frac{1}{48} \left(x^{4}-9 x^{2}\right) \cos x+\frac{1}{12} x^{3} \sin x
Example:12. \left(D^{2}+3 D+2\right) y=x^{2} \cos x
Solution: \left(D^{2}+3 D+2\right) y=x^{2} \cos x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+3 m+2=0 \Rightarrow(m+2)(m+1)=0 \\ \Rightarrow m=-1,-2

C.F.=C_{1} e^{-x}+C_{2} e^{-2 x}

P.I.=\frac{1}{(D+1)(D+2)} x^{2} \cos x \\ =\left[\frac{1}{D+1}-\frac{1}{D+2}\right] x^{2} \cos x \\ =\frac{1}{(D+1)} x^{2} \cos x-\frac{1}{D+2} x^{2} \cos x \\ =x^{2} \frac{1}{(D+1)} \cos x+2 x\left[\frac{d}{d D} \cdot \frac{1}{D+1}\right] \cos x +\left[\frac{d^{2}}{d D^{2}} \cdot \frac{1}{(D+1)}\right] \cos x-x^{2} \frac{1}{(D+2)} \cos x -2 x\left[\frac{d}{d D} \cdot \frac{1}{D+2}\right] \cos x-\left[\frac{d^{2}}{d D^{2}} \cdot \frac{1}{D+2}\right] \cos x\\ =x^{2} \frac{D-1}{D^{2}-1} \cos x-2 x \frac{1}{(D+1)^{2}} \cos x +\frac{2}{(D+1)^{3}} \cos x-x^{2} \frac{D-2}{D^{2}-4} \cos x +2 x \frac{1}{(D+2)^{2}} \cos x-\frac{2}{(D+2)^{3}} \cos x\\ =x^{2} \frac{(-\cos x-\sin x)}{i^{2}-1}-\frac{2 x}{D^{2}+2 D+1} \cos x +\frac{2}{D^{3}+3 D^{2}+3 D+1} \cos x-x^{2} \frac{(-\sin x-2 \cos x)}{i^{2}-4}+ 2 x \frac{2 x}{D^{2}+2 D+1} \cos x -\frac{2}{D^{3}+6 D^{2}+12 D+8} \cos x \\ = \frac{x^{2}}{2}(\cos x+\sin x)-\frac{2 x}{i^{2}+2 D+1} \cos x + \frac{2}{D i^{2}+3 i^{2}+3 D+1} \cos x-\frac{x^{2}}{5}(\sin x+2 \cos x+ 2 x \frac{1}{i^{2}+4 D+4} \cos x-\frac{2}{D i^{2}+6 i^{2}+12 D+8} \cos x\\ = \frac{x^{2}}{2} (\cos x+\sin x)-\frac{1}{D} \cos x+\frac{2}{-D-3+3 D+1} \cos x-\frac{x^{2}}{5} (\sin x+2 \cos x) +2 x \cdot \frac{1}{4 D-3} \cos x-\frac{2}{-D-6+12 D+8} \cos x\\ =\frac{x^{2}}{2}(\cos x+\sin x)-x \sin x \frac{D+1}{D^{2}-1} \cos x -\frac{x^{2}}{5}(\sin x+2 \cos x)+2 x \frac{4D+3}{16 D^{2}-9} \cos x-\frac{2}{11 D+2} \cos x \\ =\frac{x^{2}}{2}(\cos x+\sin x)-x \sin x+\frac{(\cos x-\sin x)}{i^{2}-1}-\frac{x^{2}}{5}(\sin x+2 \cos x)+2 x \frac{3 \cos x -4 \sin x}{16 i^{2}-9} -\frac{2(11 D+2)}{121 D^{2}-4} \cos x\\ =\frac{x^{2}}{2}(\cos x+\sin x)-x \sin x-\frac{1}{2}(\cos x-\sin x) -\frac{x^{2}}{5}(\sin x+2 \cos x) \frac{-2 x}{25}(3 \cos x-4 \sin x)-\frac{2(-11 \sin x+2 \cos x)}{-121 i^{2}-4}\\ =\frac{x^{2}}{2}(\sin x+\cos x)-x \sin x-\frac{1}{2}(\cos x-\sin x)-\frac{x^{2}}{5}(\sin x+2 \cos x) -\frac{2 x}{25}(3 \cos x-4 \sin x)-\frac{2}{125}(11 \sin x-2 \cos x) \\ \Rightarrow \text{P.I.}=\frac{x^{2}}{10}(3 \sin x+\cos x)+\frac{x}{25}(6 \cos x-17 \sin x)+\frac{1}{250}(81 \sin x-133 \cos x)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

\Rightarrow y=C_{1} e^{x}+C_{2} e^{-2 x}+\frac{x^{2}}{10}(3 \sin x+\cos x)+\frac{x}{25}(6 \cos x-17 \sin x)+\frac{1}{250}(81 \sin x-133 \cos x)
Example:13. \left(D^{2}-1\right) y=x \sin x+\left(1+x^{2}\right) e^{x}
Solution: \left(D^{2}-1\right) y=x \sin x+\left(1+x^{2}\right) e^{x}
इसका सहायक समीकरण (A.E.) होगा:

m^{2}-1=0 \Rightarrow m=\pm 1

C.F.=C_{1} e^{x}+C_{2} e^{-x}

P.I.=\frac{1}{\left(D^{2}-1\right)}\left[x \sin x+\left(1+x^{2}\right) e^{x}\right] \\ =\frac{1}{\left(D^{2}-1\right)} x \sin x+\frac{1}{\left(D^{2}-1\right)}\left(1+x^{2}\right) e^{x} \\ =\frac{1}{2} \left[\frac{1}{D-1}-\frac{1}{D+1}\right] x \sin x+e^{x} \frac{1}{(D+1)^{2}-1}\left(1+x^{2}\right) \\ =\frac{1}{2} \frac{1}{D-1}(x \sin x)-\frac{1}{2} \frac{1}{D+1} x \sin x +e^{x} \frac{1}{D^{2}+2 D+1-1}\left(1+ x^{2}\right) \\ =\frac{1}{2} x \frac{1}{D-1} \sin x+\frac{1}{2}\left[\frac{d}{d D} \cdot \frac{1}{D-1}\right] \sin x-\frac{1}{2} x \frac{1}{(D+1)} \sin x-\frac{1}{2}\left[\frac{d}{d D} \cdot \frac{1}{D+1}\right] \sin x +e^{x} \cdot \frac{1}{2 D\left(1+\frac{D}{2}\right)}\left(1+x^{2}\right)\\ =\frac{1}{2} x \frac{D+1}{D^{2}-1} \sin x-\frac{1}{2} \frac{1}{(D-1)^{2}} \sin x-\frac{1}{2} x \frac{D-1}{D^{2}-1} \sin x+\frac{1}{2} \cdot \frac{1}{(D+1)^{2}} \sin x+e^{x} \cdot \frac{1}{2 D}\left(1+\frac{D}{2}\right)^{-1} \left(1+x^{2}\right)\\ =\frac{1}{2} x \frac{(\cos x+\sin x)}{i^{2}-1}-\frac{1}{2} \frac{x}{D^{2}-2 D+1} \sin x-\frac{1}{2} x\frac{\left(\cos x-\sin x\right)}{i^{2}-1}+\frac{1}{2} \cdot \frac{1}{D^{2}+2 D+1} \sin x+e^{x}\left(\frac{1}{2 D}\right)\left(1-\frac{D}{2}+\frac{D^{2}}{4}-\frac{D^{3}}{8}+\cdots\right)(1+x^{2})\\ =-\frac{1}{4} x(\cos x+\sin x)-\frac{1}{2} \cdot \frac{1}{i^{2}-2D+1} \sin x+\frac{1}{4} x(\cos x-\sin x)+\frac{1}{2} \cdot \frac{1}{i^{2}+2 D+1} \sin x+\frac{e^{x}}{2}\left(\frac{1}{D}-\frac{1}{2}+\frac{D}{4}-\frac{D^{2}}{8}+\cdots\right)\left(1+x^{2}\right) \\ =-\frac{1}{2} \sin x+\frac{1}{4} \cdot \frac{1}{D} \sin x+\frac{1}{4} \cdot \frac{1}{D} \sin x+\frac{e^{x}}{2}\left(x+\frac{x^{3}}{3}-\frac{1}{2}-\frac{x^{2}}{2}+\frac{x}{2}-\frac{1}{4}\right)\\ =-\frac{x}{2} \sin x-\frac{1}{2} \cos x+\frac{e^{x}}{2}\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}+\frac{3}{2} x-\frac{3}{4}\right)\\ \Rightarrow \text{P.I.}=-\frac{x}{2} \sin x-\frac{1}{2} \cos x+\frac{e^{x}}{2}\left(\frac{x^{3}}{3}-\frac{x^{2}}{2}+\frac{3}{2} x-\frac{3}{4}\right)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

y=C_{1} e^{x}+C_{2} e^{x}-\frac{1}{2}(x \sin x+\cos x)+ \left(\frac{1}{12}\right) x e^{x}\left(2 x^{2}-3 x+9\right)
Example:14. \frac{d^{2} y}{d x^{2}}+y=x^{2} \sin x
Solution: \frac{d^{2} y}{d x^{2}}+y=x^{2} \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}+1=0 \Rightarrow m=\pm i

C.F.=C_{1} \cos x+C_{2} \sin x

P.I.=\frac{1}{D^{2}+1} x^{2} \sin x\\ =\text {I. P. of } \frac{1}{D^{2}+1} x^{2} e^{ix}\\ =\text {I. P. of } e^{i x} \frac{1}{(D+i)^{2}+1} x^{2}\\ =\text {I. P. of } e^{i x} \frac{1}{D^{2}+2 D i+i^{2}+1} x^{2}\\ =\text {I. P. of } e^{i x} \frac{1}{2 D i\left(1+\frac{D}{2 i}\right)} x^{2} \\ =\text {I. P. of } e^{i x}\left(\frac{-i}{2 D}\right) \left[1+\frac{D}{2 i}\right]^{-1} x^{2}\\ =\text {I. P. of } e^{i x}\left(-\frac{i}{2 D}\right)\left[1-\frac{D}{2 i}-\frac{D^{2}}{4}+\frac{D^{3}}{8 i} \cdots \right] x^{2}\\ = \text {I. P. of } e^{i x} \left(-\frac{1}{2}\right)\left[\frac{i}{D}-\frac{1}{2}-\frac{D i}{4}+\frac{D^{2}}{4}-\right] x^{2}\\ =\text {I. P. of } \left(-\frac{1}{2}\right)(\cos x+i \sin x)\left(\frac{i x^{3}}{3}-\frac{x^{2}}{2}-\frac{x i}{2}+\frac{1}{2}\right)\\ =\left(\frac{-1}{2}\right) \text {I. P. of } [\left(\frac{-x^{2}}{2} \cos x+\frac{1}{2} \cos x-\frac{x^{3}}{3} \sin x+\frac{x}{2} \sin x\right)+i\left(\frac{x^{3}}{3} \cos x-\frac{x}{2} \cos x-\frac{x^{2}}{2} \sin x+\frac{1}{2} \sin x\right)]\\ \text{P.I.}=-\frac{1}{2}\left(\frac{x^{3}}{3} \cos x-\frac{x}{2} \cos x-\frac{x^{2}}{2} \sin x+\frac{1}{2} \sin x\right)
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

y=C_{1} \cos x+C_{2} \sin x-\left(\frac{1}{12}\right)\left[\left(2 x^{3}-3 x\right) \cos x-3 x^{2} \sin x\right]
Example:15. \left(D^{2}-1\right) y=x^{2} \sin x
Solution: \left(D^{2}-1\right) y=x^{2} \sin x
इसका सहायक समीकरण (A.E.) होगा:

m^{2}-1=0 \\ \Rightarrow(m-1)(m+1)=0\\ \Rightarrow m=1,-1

C.F.=C_{1} e^{x}+C_{2} e^{-x}

P.I.=\frac{1}{D^{2}-1} x^{2} \sin x =\text{I. P. of } \frac{1}{D^{2}-1} x^{2} e^{ix}\\ =\text{I. P. of } e^{i x} \frac{1}{(D+i)^{2}-1} x^{2}\\ =\text{I. P. of } e^{i x} \frac{1}{D^{2}+2 D i+i^{2}-1} x^{2}\\ =\text{I. P. of } e^{i x} \frac{1}{D^{2}+2 D i-2} x^{2}\\=\text{I. P. of } \left(-\frac{1}{2} e^{ix}\right) \left [ 1-\left(\frac{1}{2} D^{2}-D i\right)\right]^{-1} x^{2}\\ =\text{I. P. of } \left(-\frac{1}{2} e^{i x}\right)\left[1+\frac{1}{2} D^{2}+D i+\left(\frac{1}{2} D^{2}+D i\right)^{2}+\cdots\right] x^{2}\\ =\text{I. P. of } \left(-\frac{1}{2} e^{i x}\right)\left[1+\frac{1}{2} D^{2} +D i-D^{2}+ \cdots\right] x^{2}\\ =\text{I. P. of } \left(-\frac{1}{2}\right)(\cos x+i \sin x)\left(x^{2}-1+2 x i\right)\\ =\text{I. P. of } (-\frac{1}{2})\left[\left(x^{2}-1\right) \cos x-2 x \sin x+i \left(2 x \cos x+x^{2} \sin x-\sin x\right)\right]
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

y=C_{1} e^{x}+C_{2} e^{-x}-x \cos x-\frac{1}{2}\left(x^{2}-1\right) \sin x
Example:16. \left(D^{2}+1\right)^{2} y=24 x \cos x
जब (When) x=0 तो (then) y=0

D y=0, D^{2} y=0, D^{2}y=12
Solution: \left(D^{2}+1\right)^{2} y=0
इसका सहायक समीकरण (A.E.) होगा:

\left(m^{2}+1\right)^{2}=0, \Rightarrow m=\pm i, \pm i

C.F.=(C_{1} x+C_{2}) \cos x+\left(C_{3} x+C_{4}\right) \sin x

P.I.=\frac{1}{\left(D^{2}+1\right)^{2}} 24 \cos x\\ =\text{R.P. of } \frac{1}{\left(D^{2}+1\right)^{2}} 24 x e^{i x}\\ =\text{R.P. of } 24 e^{i x} \frac{1}{\left[(D+i)^{2}+1\right]^{2}} x \\ =\text{R.P. of } 24 e^{i x} \frac{1}{\left[D^{2}+2 D i+i^{2}+1\right]^{2}} x\\ =\text{R.P. of } 24 e^{i x} \frac{1}{\left(D^{2}+2 D i\right)^{2}}x \\ =\text{R.P. of } 24 e^{i x} \frac{1}{4 D^{2} i^{2}\left(1+\frac{D}{2 i}\right)^{2}} \\= \text{R.P. of } 24 e^{i x}\left(\frac{1}{-4 D^{2}}\right)\left(1+\frac{D}{2 i}\right)^{-2} x \\ =\text{R.P. of } \left(-6 e^{i x}\right) \frac{1}{D^{2}}\left[1+i D-\frac{3}{4} D^{2}+ \cdots \right]x \\=\text{R.P. of }\left(-6 e^{i x}\right) \cdot \frac{1}{D^{2}}(x+i) \\ = \text{R.P. of } (-6)(\cos x+i \sin x)\left(\frac{x^{3}}{6}+\frac{i x^{2}}{2}+\cdots \right) \\ \text{P.I.}=-x^{2} \cos x+3 x^{2} \sin x
अतः दिए हुए समीकरण का व्यापक हल होगा:
y=C.F.+P.I.

y=(C_{1} x+C_{2}) \cos x+\left(C_{3} x+C_{4}\right) \sin x -x^{2} \cos x+3 x^{2} \sin x \\ \Rightarrow y=\left(C_{1} x+C_{2}-x^{2}\right) \cos x+\left(C_{2} x+C_{4}+3 x^{2}\right) \sin x \cdots(1) \\ D y =\left(C_{1} -3 x^{2}\right) \cos x-\left(C_{1} x+C_{2}-x^{2}\right) \sin x +\left(C_{3} x+C_{4}+3 x^{2}\right) \cos x+(C_{3}+6 x) \sin x \\D y=\left(C_{1}+C_{4}+C_{3} x\right) \cos x+\left(C_{3}-C_{2}+6 x-C_{1} x+x^{3}\right) \sin x \cdots(2) \\ D^{2} y=-\left(C_{1}+C_{4}+C_{3} x\right) \sin x+C_{3} \cos x+C_{3}-C_{2}+6 x-C_{1} x +x^{3}) \cos x+\left(6-C_{1}+3 x^{2}\right) \sin x \\ D^{2} y=\left(6-2 C_{1}-C_{4}-C_{3} x+3 x^{2}\right) \sin x+\left(2 C_{3}-C_{2}+(x-C_{1}x+ x^{3}\right) \cos x \cdots(3) \\ D^{3} y=\left(6-2 C_{1}-C_{4}-C_{3} x+3 x^{2}\right) \cos x+\left (-C_{3}+6x \right )\sin x-\left(2C_{3}-C_{2}+6 x-C_{1} x+x^{3}\right) \sin x+\left(6-C_{1}+3 x^{2}\right) \cos x \\ D^{3} y=\left(12-3 C_{1}-C_{4}-C_{3} x+6 x^{2}\right) \cos x+\left(C_{2}-3 C_{3}-6 x+C_{1} x+6 x-x^{2}\right) \sin x \cdots(4)
जब x=0,y=0 तो D y=0, D^{2} y=0, D^{3} y=12
अतः समीकरण (1),(2),(3) तथा (4) से:

0=C_{1} \cdots(5) \\ 0=C_{1}+c_{4} \cdots(6) \\ 0=2 C_{3}+C_{2} \cdots(7) \\ 12=12-C_{1}-C_{4} \cdots(8)
समीकरण (5) तथा (7) से:

C_{1}=0=C_{4}
समीकरण (6) तथा (8) से:

C_{2}=0=C_{4}
उपर्युक्त मान (1) में रखने पर:

y=3 x^{2} \sin x-x^{3} \cos x
उपर्युक्त उदाहरणों के द्वारा अवकल समीकरण का विशिष्ट समाकल (Particular Integral of DE),पूरक फलन और विशिष्ट समाकल (Complementary Function and Particular Integral) को समझ सकते हैं।

3.अवकल समीकरण का विशिष्ट समाकल पर आधारित सवाल(Questions Based on Particular Integral of DE Solved Examples):

(1.) \left(D^{4}-1\right) y=x^{2} \sin x \\ (2.) \left(D^{2}+m^{2}\right) y=x \cos m x

उत्तर (Answers): (1.) y=C_{1} e^{x}+C_{2} e^{-x}+C_{3} \cos x+C_{4} \sin x+\frac{1}{12} x^{3} \cos x-\frac{5}{8} x \cos x-\frac{3}{8} x^{2} \sin x

(2.) y=C_{1} \cos m x+C_{2} \sin mx +\frac{x^{2}}{4 m} \sin mx
उपर्युक्त सवालों को हल करने पर अवकल समीकरण का विशिष्ट समाकल (Particular Integral of DE),पूरक फलन और विशिष्ट समाकल (Complementary Function and Particular Integral) को ठीक से समझ सकते हैं।

Also Read This Article:-Particular Integral in Special Cases

4.अवकल समीकरण का विशिष्ट समाकल (Particular Integral of DE),पूरक फलन और विशिष्ट समाकल (Complementary Function and Particular Integral) के सम्बन्ध में अक्सर पूछे जाने वाले प्रश्न:

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Particular Integral of DE

अवकल समीकरण का विशिष्ट समाकल
(Particular Integral of DE)

Particular Integral of DE