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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

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Answer:

Explanation

Standard form


y

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

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


e^x V because
e^x\\e 0

Let U = V'


U'+5U=0\rightarrow (du)/(u)=-5dx .


lnu= -5x \rightarrow u = e^(-5x)

Replace back,


V' = e^(-5x)\rightarrow V= (e^(-5x))/(-5)

then


y(x) = V y_1(x) = (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

User Vortico
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