Question

solve the differential equation by variation of parameters, subject to the initial conditions y(0) = 1, y'(0) = 0. 9y'' − y = xe^x/3.

Answer 1

`y=(1)/(8)e^{-(x)/(3)}+(7)/(8)e^{(x)/(3)}-(1)/(4)xe^{(x)/(3)}+(1)/(12)x^(2)e^{(x)/(3)}`

Explanation:

Explanation is attached below. Please examine in chronological order.

Answer 2

The given differential equation is:

9y" - y = xe^(x/3)

The associated homogeneous equation is:

9y" - y = 0

The characteristic equation is:

9r^2 - 1 = 0

Solving for r, we get:

r = ±1/3

So, the general solution of the homogeneous equation is:

y_h(x) = c1e^(x/3) + c2e^(-x/3)

Now, we need to find the particular solution of the non-homogeneous equation using variation of parameters.

Let the particular solution be:

y_p(x) = u1(x)e^(x/3)

y_p' = u1'e^(x/3) + u1/3 * e^(x/3)

y_p" = u1"e^(x/3) + 2u1'/3 * e^(x/3) + u1/9 * e^(x/3)

Substituting these into the differential equation, we get:

9(u1"e^(x/3) + 2u1'/3 * e^(x/3) + u1/9 * e^(x/3)) - u1e^(x/3) = xe^(x/3)

Simplifying, we get:

u1"e^(x/3) + 2u1'/3 * e^(x/3) = x/81

Multiplying both sides by e^(2x/3), we get:

(u1'e^(x/3))^' = x/81 * e^(2x/3)

Integrating both sides, we get:

u1'e^(x/3) = (27/4) * e^(2x/3) - (9/2) * x * e^(2x/3) + C1

where C1 is the constant of integration.

Integrating both sides again, we get:

u1(x) = (27/4) * e^(x/3) - (27/4) * x * e^(x/3) + (9/4) * x^2 * e^(x/3) + C1 * e^(-x/3) + C2

where C2 is the constant of integration.

Therefore, the general solution of the non-homogeneous equation is:

y(x) = y_h(x) + y_p(x) = c