Question

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation and a particular solution yp(x) of the given nonhomogeneous equation.y'' − 3y' + 2y = 7e3x; y1 = ex

Answer 1

Answer:

Step-by-step explanation

Standard form

$y" + P(x)y'+Q(x) y =0$

Hence your P(t) = -3, Q(t) = 2

$y(x) = Vy_1(x)\rightarrow y'(x) = V'y_1(x) + Vy_1'(x) ~~~and ~~~y"(x) = v"y_1(x) +2V'y_1'(x)+Vy_1(x)$

After replacing all y", y' and y to homogeneous part, you will have

$e^x V"+5e^x V'=0\longleftrightarrow V"+5V'=0$ because $e^x\ne 0$

Let U = V'

$U'+5U=0\rightarrow \frac{du}{u}=-5dx$ .

$lnu= -5x \rightarrow u = e^{-5x}$

Replace back,

$V' = e^{-5x}\rightarrow V= \frac{e^{-5x}}{-5}$

then

$y(x) = V y_1(x) = \frac{e^{-4x}}{-5}$

So, general solution of the ODE is

$Ay_1(x) + By_2(x)$

Particular solution is just take derivative of the general one twice and plug back into the original ODE to find A and B

You can finish it by yourself. Let me know if you need more help