Question

Solve d2y/dx2-4y=ex+sin2x​

Answer 1

Answer:

Your answer

this is step by step explained...

Answer 2

Answer:

Solution of $\dfrac{d^2y}{dx^2} -4y=e^x+sin2x$ is $C_{1}e^2x+ C_{2}e^{-2x}-\dfrac{1}{3} e^x- \dfrac{x}{4}cos2x$

Step-by-step explanation:

Given an inhomogeneous equation,

$\dfrac{d^2y}{dx^2} -4y=e^x+sin2x$

Using the differential operator, the left hand side can be written as

$(D^2 -4)y=e^x+sin2x$

The auxiliary equation is $D^2 -4=0$

$\implies D^2 =4\implies D =\pm2$

Hence the complementary function is $y(CF)=C_{1}e^2x+ C_{2}e^{-2x}$

Now for particular Integral,

$y(PI)=\dfrac{1}{D^2-4} e^x+ sin2x$

$=\dfrac{1}{D^2-4} e^x+ \dfrac{1}{D^2-4}sin2x$

$=\dfrac{1}{1^2-4} e^x+ \dfrac{1}{(-2)^2-4}sin2x$

$=-\dfrac{1}{3} e^x+ \dfrac{1}{4-4}sin2x$

Since the denominator in the second term is zero,

$y(PI)=-\dfrac{1}{3} e^x- \dfrac{x}{4}cos2x$

Hence the required solution is

$y=y(CF)+y(PI)$

$=C_{1}e^2x+ C_{2}e^{-2x}-\dfrac{1}{3} e^x- \dfrac{x}{4}cos2x$

Solution of the given equation is $C_{1}e^2x+ C_{2}e^{-2x}-\dfrac{1}{3} e^x- \dfrac{x}{4}cos2x$