Question

sove it :y'' +4y = e^2x​

Answer 1

may be it's wrong but I try.....

Answer 2

Answer:

here i answered your question

Step-by-step explanation:

the general solution of the homogeneous equation

$y^{n}-4y =0$

is given by theorem 8.7 with a=0 and b=-4this give us d=a²-4b=16

hence k=$\frac{1}{2}\sqrt{d} =2$

$y= e^{-\frac{ax}{2} } (c_{1} e^{kx} +c_{2} e^{-kx}   )$

$= c_{1}e^{2x}  + c_{2}e^{-2x}$

to find a particular solution of

$y^{n}-4y=c^{2x}$

assume  

$y_{1} = p(x)e^{2x}$

is a solution. then,

$y'_{1} = 2p(x)e^{2x} + p'(x)e^{2x}</p><p> y_{1}^{n}  = p^{n}(x)e^{2x} + 4p'(x)e^{2x}  + 4p(x)e^{2x}$

therefore,

⇒

$p^{n}(x)e^{2x} + 4p'('x)e^{2x} + 4p(x)e^{2x} - 4p(x)e^{2x} = e^{2x}$

⇒

$p^{n} (x) + 4p'(x) = 1$

now, let p(x)= Ax+B, then we have

4A=1

⇒

A=1/4

p(x)= 1/4 x

thus,

$y_1  = \frac{1}{4} xe^{2x}.$

hence the general solution given by

$y=c_1e^{2x} + c^{2}e^{-2x} + \frac{1}{4} x e^{2x} = e^{2x}(\frac{1}{4}x + c_1)  + c_2e^{-2x}$

hope it helps you