Math, asked by Shilpa00, 6 months ago

Solve the Diffrential equation

d²y/dx² + dy/dx + y = Cos 2x
Include complete solution. [ Complementary Function + Particular Integral ]​

Answers

Answered by BrainlyPopularman
31

GIVEN :–

• A differential equation –

 \\ \implies \bf  \dfrac{ {d}^{2}y }{ {dx}^{2} }  + \dfrac{dy}{dx}  + y =  \cos(2x)  \\

TO  FIND :–

• Solution of differential = ?

SOLUTION :–

• Solution of given differential equation –

 \\ \implies \bf  Solution =Complementary \:  \: Function + Particular \:  \: Integral \\

 \\ \implies\large{ \boxed {\bf  Solution =C.F.+ P. I.}} \\

• Let find C.F. –

 \\ \implies \bf  \dfrac{ {d}^{2}y }{ {dx}^{2} }  + \dfrac{dy}{dx}  + y = 0  \\

• We should write this as –

 \\ \implies \bf ({D}^{2} +D+1)y = 0  \\

• Auxiliary equation –

 \\ \implies \bf {m}^{2} +m+1= 0  \\

 \\ \implies \bf m =  \dfrac{ - 1 \pm \sqrt{ {(1)}^{2} - 4(1)(1) } }{2(1)}\\

 \\ \implies \bf m =  \dfrac{ - 1 \pm \sqrt{1- 4} }{2(1)}\\

 \\ \implies \bf m =  \dfrac{ - 1 \pm \sqrt{ - 3} }{2}\\

 \\ \implies \bf m =  \dfrac{ - 1 \pm \sqrt{3}i}{2}\\

• Hence , C.F. –

 \\ \implies \large{ \boxed{ \bf C.F. =  {e}^{ - \frac{1}{2}x } \bigg[c_1 \cos \bigg( \dfrac{ \sqrt{3}x }{2} \bigg) + c_2\sin \bigg( \dfrac{ \sqrt{3}x }{2} \bigg)\bigg] }}\\

• Now let's find P.I. –

 \\ \implies \bf P.I. =  \dfrac{1}{F(D)}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{1}{{D}^{2} +D+1}  \cos(2x) \\

• We know that –

 \\ \implies \large{ \boxed{ \bf \dfrac{1}{F(D^{2} )}\cos(ax)= \dfrac{1}{F( - a^{2} )}\cos(ax) \: \:  ,  \: \: F( - a^{2} ) \neq0}}\\

• So that –

 \\ \implies \bf P.I. =  \dfrac{1}{ - {(2)}^{2} +D+1}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{1}{ - 4+D+1}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{1}{D-3}  \cos(2x) \\

• Now Rationalization –

 \\ \implies \bf P.I. =  \dfrac{1(D+3)}{(D-3)(D+3)}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{(D+3)}{(D^{2}  - 9)}  \cos(2x) \\

• Again using identity –

 \\ \implies \bf P.I. =  \dfrac{(D+3)}{ \{ - (2)^{2}  - 9 \}}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{(D+3)}{ ( - 4  - 9 )}  \cos(2x) \\

 \\ \implies \bf P.I. =  \dfrac{(-3 - D)}{13}  \cos(2x) \\

 \\ \implies \bf P.I. = -\dfrac{3}{13}  \cos(2x) - \dfrac{D}{13}  \cos(2x) \\

 \\ \large\implies{ \boxed{\bf P.I. =-  \dfrac{3}{13}  \cos(2x)+\dfrac{2}{13}  \sin(2x)}}\\

 \\ \large\implies{ \boxed{\bf P.I. =\dfrac{2}{13}  \sin(2x)-\dfrac{3}{13}  \cos(2x)}}\\

• Hence , Solution –

\bf\implies{\boxed {\bf {Solution ={e}^{ - \frac{1}{2}x } \bigg[{c}_{1} \cos \bigg( \dfrac{ \sqrt{3}x }{2} \bigg) + {c}_{2} \sin \bigg( \dfrac{ \sqrt{3}x }{2} \bigg)\bigg ]+ \dfrac{2}{13}  \sin(2x)-\dfrac{3}{13}  \cos(2x)}} }


pulakmath007: Excellent
BrainlyPopularman: Thank you sir
Answered by hukam0685
7

Step-by-step-Explanation

Given:

 \frac{ {d}^{2}y }{d {x}^{2} }  +  \frac{dy}{dx}  + y = cos2x \\

To find:Solve the Differential equation

Solution:

Solution=C.F.+P.I.

To solve the differential equation first write the auxiliary equation.

 {m}^{2}  + m + 1 = 0 \\

It can't be factorize,so applying quadratic formula

m =  \frac{ - b ±  \sqrt{ {b}^{2}  - 4ac} }{2a}  \\  \\ here \\  \\ a = 1 \\ b = 1 \\ c = 1 \\  \\ m = \frac{ - 1 ±\sqrt{ {( - 1)}^{2}  - 4} }{2} \\  \\ m =  \frac{ - 1 ± i \sqrt{3} }{2}  \\  \\ m_1 =  \frac{ - 1 - i \sqrt{3} }{2}  \\  \\ m_2 = \frac{ - 1  +  i \sqrt{3} }{2} \\  \\

Hence,

CF

y = c_1 {e}^{m_1x}  + c_2 {e}^{m_2x}  \\  \\

y=c_1{e}^{x(\frac{-1-i\sqrt{3}}{2})}+c_2{e}^{x(\frac{-1+i\sqrt{3}}{2})}\\  \\

Now,

Find Particular Integral,

for that assume equation to be non-homogeneous

~y=Acos2x+Bsin2x\\

put the value in

 \frac{ {d}^{2}y }{d {x}^{2} }  +  \frac{dy}{dx}  + y = cos2x \\

and solve for A and B

 \frac{ {d}^{2}(Acos2x+Bsin2x) }{d {x}^{2} }  +  \frac{d(Acos2x+Bsin2x)}{dx}  + (Acos2x+Bsin2x) = cos2x \\\\-4Acos2x-4Bsin2x-2Asin2x+2Bcos2x+Acos2x+Bsin2x=cos2x\\\\

sin2x and cos2x are linearly independent,compare the coefficient in LHS and RHS

-4A+2B+A=1

or

-3A+2B=1...eq1

-2A-4B+B=0

or

-2A-3B=0...eq2

Solve eq1 and 2 to find value of A and B

- 6A+4B=2

-6A-9B=0

(+) (+) (-)

_________

13B=2

B=2/13

-2A-3×2/13=0

-2A=6/13

A=-3/13

Thus,

Solution of differential equation is

y=y_{homogeneous}+y_{non-homogeneous}\\\\y=c_1{e}^{x(\frac{-1-i\sqrt{3}}{2})}+c_2{e}^{x(\frac{-1+i\sqrt{3}}{2})}-\frac{3}{13}cos2x+\frac{2}{13}sin2x\\\\

Thus,

This is the completed Solution of given differential equation.

Hope it helps you.

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