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#### karush

##### Well-known member

- Jan 31, 2012

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Find the general solution of the given differential equation

$\displaystyle ty^\prime -y =t^2e^{-1}\\$

Divide thru by t

$\displaystyle y^\prime-\frac{y}{t}=te^{-1}\\$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -\frac{1}{t} \, dt =\frac{1}{t}\\$

Multiply thru with $\displaystyle\frac{1}{t}$

$\displaystyle y^\prime -\frac{1}{t}y =te^{-1}\\$

Simplify:

$(\frac{1}{t}y)'=te^{-1} \\$

ok got stuck here so...

Answer from bk

$\displaystyle

\color{red}{y=c_1 -te^{-1}}$ (this is wrong)

$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value disappearProblems}$$

$\displaystyle ty^\prime -y =t^2e^{-1}\\$

Divide thru by t

$\displaystyle y^\prime-\frac{y}{t}=te^{-1}\\$

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -\frac{1}{t} \, dt =\frac{1}{t}\\$

Multiply thru with $\displaystyle\frac{1}{t}$

$\displaystyle y^\prime -\frac{1}{t}y =te^{-1}\\$

Simplify:

$(\frac{1}{t}y)'=te^{-1} \\$

ok got stuck here so...

Answer from bk

$\displaystyle

\color{red}{y=c_1 -te^{-1}}$ (this is wrong)

$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value disappearProblems}$$

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