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{\bf Question}

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Find the solution of the initial value problem $$ \frac{dx}{dt} +
\frac{3}{t}x = t - \frac{2}{t^2}$$

 given $x(1) = 1$

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{\bf Answer}

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$$\frac{dx}{dt} + \frac{3}{t}x = t - \frac{2}{t^2}$$

with $x(1) = 1$ for $ t>1$

Here the integrating factor is $\displaystyle  e^{\int \frac{3}{t}
\, dt} = e^{3 ln|t|} = |t|^3 = t^3$ for $t>1$



Thus $\displaystyle t^3 \frac{dx}{dt} + 3t^2x = t^4 - 2t$
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Hence  $\displaystyle \frac{d}{dt}(t^3x)  =  t^4 - 2t \Rightarrow
xt^3  =  \frac{1}{5}t^2 - \frac{1}{t} + \frac{C}{t}$

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Where  $C$ is a constant.

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Find $C$ using initial condition $\displaystyle 1  =  \frac{1}{5}
- 1 + C \Rightarrow C = \frac{9}{5}  \Rightarrow x  =
\frac{1}{5}t^2 - \frac{1}{t} + \frac{9}{5t^3}$



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