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karush

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Find the general solution of the given differential equation

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

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -2t \, dt =e^{-t^2}$$%e^(-t^2)\\$

Multiply thru with $e^{-t^2}$

$(e^{-t^2})y^\prime -(e^{-t^2})2ty =(e^{-t^2})2te^{-t^2}\\$

Simplify:

$((e^{-t^2})y)'= 2te^{-2t^2}\\$

Integrate:

$\displaystyle e^{-t^2}y=\int 2te^{-2t^2} dt =-\frac{ e^{-2t^2}}{2}+c_1\\$

Divide by $e^{-t^2}$

$\displaystyle y=-\frac{e^{-t^2}}{2}+c_1 e^{t^2}\\$

Answer from $\textbf{W|A}$

$\displaystyle y=\color{red}{c_1 e^{t^2}-\frac{e^{-t^2}}{2}}$

ok got to be some typos in this

otherwise suggestions$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value Problems}$$

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

Obtain $u(t)$

$\displaystyle u(t)=\exp\int -2t \, dt =e^{-t^2}$$%e^(-t^2)\\$

Multiply thru with $e^{-t^2}$

$(e^{-t^2})y^\prime -(e^{-t^2})2ty =(e^{-t^2})2te^{-t^2}\\$

Simplify:

$((e^{-t^2})y)'= 2te^{-2t^2}\\$

Integrate:

$\displaystyle e^{-t^2}y=\int 2te^{-2t^2} dt =-\frac{ e^{-2t^2}}{2}+c_1\\$

Divide by $e^{-t^2}$

$\displaystyle y=-\frac{e^{-t^2}}{2}+c_1 e^{t^2}\\$

Answer from $\textbf{W|A}$

$\displaystyle y=\color{red}{c_1 e^{t^2}-\frac{e^{-t^2}}{2}}$

ok got to be some typos in this

otherwise suggestions$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value Problems}$$

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