Answer to Question #109293 in Differential Equations for sudhir

Question

Solve the following initial value problem

d^2 y/dx^2 + dy/dx -2y = -6 sin2x - 18 cos2x
y(0)=2 , y'(0)=2

Answer 1

The given differential equation is

$$\frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=-6sin(2x)-18cos(2x)$$

The auxiliary equation is $m^2+m-2=0.$

$\therefore \ m^2+m-2=0$

$\implies m^2+2m-m-2=0$

$\implies m(m+2)-1(m+2)=0$

$\implies (m+2)(m-1)=0$

$\implies m=1,-2$ .

$\therefore \ C.F.=Ae^x+Be^{-2x}$

Where ,$C.F.=$ Complementary Function.

Now ,

$$P.I.=\frac{1}{D^2+D-2}(-6sin2x-18cos2x)$$

$$=\frac{1}{D^2+D-2}(-6sin2x)+\frac{1}{D^2+D-2}(-18cos2x)$$ $$=-6×\frac{1}{-2^2+D-2}sin2x-18×\frac{1}{-2^2+D-2}cos2x$$ $$=-6×\frac{1}{D-6}sin2x-18×\frac{1}{D-6}cos2x$$ $$=-6×\frac{D+6}{D^2-36}sin2x-18×\frac{D+6}{D^2-36}cos2x$$

Multiplying numerator and denomination by $D+6$ .

$$=-6×\frac{D+6}{-4-36}sin2x-18×\frac{D+6}{-4-36}cos2x$$

$=\frac{3}{20}×(D+6)sin2x+\frac{9}{20}×(D+6)cos2x$

$=\frac{3}{20}×(2cos2x+6sin2x)+\frac{9}{20}×(-2sin2x+6cos2x)$

$=cos2x(\frac{3}{10}+\frac{27}{10})+sin2x(\frac{9}{10}-\frac{9}{10})$

$=3cos2x$

Therefore the complete solution is

$y=C.F.+P.I$

$=Ae^x+Be^{-2x}+3cos2x..............(1)$

Again differentiate equation (1) ,we get

$y'(x)=Ae^x-2Be^{-2x}-6sin2x \ ....................(2)$

As given that ,

$y(0)=2 \ and \ y'(0)=2$

So from equation (1) and (2) ,we get

$A+B+3=2$

$\implies A+B=-1 \ ..............(3)$

and $A-2B=2....................(4)$

Subtracting (4) from (3) ,we get

$A+B-(A-2B)=-1-2$

$\implies 3B=-3 \ \implies B=-1$

Again from (3) ,we get

$A-1=-1 \ \implies A=0$ .

Therefore equation (1) become

$y=-e^{-2x}+3cos2x$ .