### Author Topic: TUT0801 Quiz 1  (Read 480 times)

#### bella

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##### TUT0801 Quiz 1
« on: September 28, 2019, 11:18:38 AM »
$$t^{3} y^{\prime}+4 t^{2} y=e^{-t}, \quad y(-1)=0, \quad t<0$$
First divide by $t^3$ on both side of the equation, we get

$$y^{\prime}+\frac{4}{t} y=\frac{e^{-t}}{t^{3}}$$

Using the method of integrating factor we have equation for  $u(t)$

$$u(t)=e^{\int \frac{4}{t} d t}=e^{4 \ln (t)} =t^{4}$$

Multiply both sides by $t^{4}$

$$t^{4} y^{\prime}+4 t^{3} y=t e^{-t}$$

$$\left(t^{4} y\right)^{\prime}=t e^{-t}$$

$$t^{4} y=\int t e^{-t}$$

$$t^{4} y=-t e^{-t}-e^{-t}+C$$

$$y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}+\frac{C}{t^{4}}$$
to check $C,$ plug in condition $y(-1)=0$

$$y(-1)=e-e+C=C=0$$
Plug in $C=0$ gets

$$y=-\frac{e^{-t}}{t^{3}}-\frac{e^{-t}}{t^{4}}$$