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[SOLVED] 2.1.10 ty' -y =t^2e^{-1} u(x)???

karush

Well-known member
Jan 31, 2012
2,928
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)

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

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
$\displaystyle
\color{red}{y=c_1 -te^{-1}}$

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

I think you have done a minor mistake; observe that; $$t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$$.

Hope you can do it from here :)
 

karush

Well-known member
Jan 31, 2012
2,928
Ok mucho mahalo
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621

karush

Well-known member
Jan 31, 2012
2,928
Hi karush, I think you have done a minor mistake; observe that;
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$. Hope you can do it from here
So if
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$
then
$\displaystyle t\left(\frac{y}{t}\right)'=te^{-1}$
or $\displaystyle
\left(\frac{y}{t}\right)'=e^{-1}$
then
$\displaystyle \frac{y}{t}=\int e^{-1} dt=\frac{t}{e}+c_1$
multiply thru by t
$\displaystyle y=\frac{t^2}{e}+c_1t$

kinda maybe hopefully raj
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
So if
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$
then
$\displaystyle t\left(\frac{y}{t}\right)'=te^{-1}$
or $\displaystyle
\left(\frac{y}{t}\right)'=e^{-1}$
then
$\displaystyle \frac{y}{t}=\int e^{-1} dt=\frac{t}{e}+c_1$
multiply thru by t
$\displaystyle y=\frac{t^2}{e}+c_1t$

kinda maybe hopefully raj
Yes that is correct. Well done. :)