MHB -2.1.10 solve ty' -y =t^2e^{-1} u(x)

[karush]

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)

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

 

[Sudharaka]

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]

Ok mucho mahalo

 

[Sudharaka]

Ok mucho mahalo

Uhmmm
You should live in a neighborhood
Where dogs bark and hal 24/7

No worries. :)

Also I note that the provided answer is wrong; it should be,

https://www.wolframalpha.com/input/?i=ty'-y=(t^2)/e

 

[karush]

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

 

[Sudharaka]

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. :)