differential equation

[Timson28]

I solved this differential equation but I'm not sure if I'm right.

Would you check it?

Thanks in advance

 

[Zermelo]

Try to plug the solution in the original equation, the same way you would verify solutions of a “normal” (for example linear) equation, see what will you get.

 

[HallsofIvy]

First note that \(\displaystyle e^{a+ b}= e^ae^b\) so you can write your solution as
\(\displaystyle y= e^{c_1}e^{-x}- 1\).

And, since \(\displaystyle c_1\) is an arbitrary constant, so is \(\displaystyle e^{c_1}\) so let \(\displaystyle C= e^{c_1}\) and
\(\displaystyle y= Ce^{-x}- 1\).

Now, \(\displaystyle y'= -Ce^{-x}\) so what is \(\displaystyle y'+ xy+ x\)?

 

[skeeter]

[MATH]\frac{dy}{dx} = -x(y+1)[/MATH]
[MATH]\frac{dy}{y+1} = -x \, dx[/MATH]
now integrate …

 

[Timson28]

omg i see my foult. Thx

 

[Deleted member 4993]

I solved this differential equation but I'm not sure if I'm right.

Would you check it?

Thanks in advance

The best way to check your answer is:

Differentiate the function you derived and see whether that satisfies the given DE.​

 

[Steven G]

I solved this differential equation but I'm not sure if I'm right.

Would you check it?

Thanks in advance

The left side has x and the right side has y'. How does that change to dx on the left and dy on the right? Try "multiplying" both sides by dx