MHB -b.2.1.5 Find the general solution of y' - 2y =3te^t,

[karush]

\nmh{895}
Find the general solution of the given differential equation
$\displaystyle y^\prime - 2y =3te^t, \\$

Obtain $u(t)$
$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t} \\$
Multiply thru with $e^{-2t}$
$e^{-2t}y^\prime + 2e^{-2t}y= 3te^{-t} \\$
Simplify:
$(e^{-2t}y)'= 3te^{-t} \\$
Integrate:
$\displaystyle e^{-2t}y=\int 3te^{-t} dt =-3e^{-t}(t+1)+c_1 \\$
{Divide by $e^{-2t}$
$y=-3e^t t - 3e^{-t} +c_1 e^t \\$
Answer from textbook
$y=\color{red}{c_1 e^{2t}-3e^t}$

ok sumtum went wrong somewhere?

$$\tiny\color{blue}{\textbf{Text book: Elementary Differential Equations and Boundary Value Problems}}$$

 

[MarkFL]

I with you up to after the integration:

$$e^{-2t}y=-3e^{-t}(t+1)+c_1$$

Thus:

$$y(t)=-3e^{t}(t+1)+c_1e^{2t}$$

The solution given by the book is incorrect.

 

[karush]

these problems require mega being carefull!