# (1-t)y” +ty’ – y=2(1-t)^(2)e^(-t), 0 Q&A By tamdoan · October 26, 2022 · 0 Comment

Use the method of variation of parameters when solving part b.

a. Verify that y_1( t)=e^t and y_2( t) = t are solutions to the homogeneous equation.

b. Find a particular solution to the differential equation.

• a) Direct substitution yields

(1 - t) e^t + t e^t - e^t = 0, and (1 - t) * 0 + t * 1 - t = 0.

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b) Assume that y = e^t * u + t * v.

Differentiating,

y' = e^t u + e^t u' + v + tv'.

Setting e^t u' + tv' = 0 (*), we obtain

y' = e^t u + v.

Differentiating again,

y'' = e^t u' + e^t u + v'.

Substitute into the DE:

(1 - t)(e^t u' + e^t u + v') + t(e^t u + v) - (e^t u + tv) = 2(1 - t)^2 e^(-t)

==> e^t u' + v' = 2(1 - t) e^(-t) (**).

From (*) and (**), we have two equations in (*) and (**):

e^t u' + tv' = 0

e^t u' + v' = 2(1 - t) e^(-t).

Solving for u' and v' yields

u' = -2te^(-2t) , and v' = 2e^(-t)

Integrating, we obtain

u = te^(-2t) + (1/2)e^(-2t) + A, and v = -2e^(-t) + B.

Hence, the general solution is

y = e^t [te^(-2t) + (1/2)e^(-2t) + A] + t [-2e^(-t) + B]

...= (Ae^t + Bt) + e^(-t) (-t + 1/2).

So, a particular solution is y = e^(-t) (-t + 1/2).

I hope this helps!