MHB 2.1.2 Find the general solution of the given differential equation

[karush]

Find the general solution of the given differential equation
$\displaystyle y^\prime - 2y = t^2 e^{2t}$

Obtain $u(t)$
$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t}$
Multiply thru with $e^{-2t}$

$e^{-2t}y^\prime
+ 2e^{-2t}y
= t^2 $

Simplify:
$(e^{-2t}y)'= t^2$

Integrate:
$\displaystyle e^{-2t}y=\int t^2\, dt=-\frac{t^3}{3}+c_1$
Divide thru by $e^{-2t}$

$\displaystyle -\frac{t^3e^{2t}}{3}+c_1e^{2t}$

ok took me 2 hours hope it ok
any suggest?

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

 

[tkhunny]

Find the general solution of the given differential equation
$\displaystyle y^\prime - 2y = t^2 e^{2t}$

Obtain $u(t)$
$\displaystyle u(t)=\exp\int -2 \, dx =e^{-2t}$
Multiply thru with $e^{-2t}$

$e^{-2t}y^\prime
+ 2e^{-2t}y
= t^2 $

Simplify:
$(e^{-2t}y)'= t^2$

Integrate:
$\displaystyle e^{-2t}y=\int t^2\, dt=-\frac{t^3}{3}+c_1$
Divide thru by $e^{-2t}$

$\displaystyle -\frac{t^3e^{2t}}{3}+c_1e^{2t}$

ok took me 2 hours hope it ok
any suggest?

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

Did you check your result? Seems as though the integral didn't quite happen correctly.

 

[karush]

Did you check your result? Seems as though the integral didn't quite happen correctly.

$$\displaystyle e^{-2t}y=\int t^2\, dt=\frac{t^3}{3}+c_1$$

you must mean the negative sign?
which I took out