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[SOLVED] 2.1.5 Find the general solution of the given differential equation

karush

Well-known member
Jan 31, 2012
2,928
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 text book
$y=\color{red}{c_1 e^{2t}-3e^t}$

ok sumtum went wrong somewhere?

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I with you up to after the integration:

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

Thus:

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

The solution given by the book is incorrect.
 

karush

Well-known member
Jan 31, 2012
2,928
these problems require mega being carefull!