differential equation

Timson28

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Sep 6, 2020
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I solved this differential equation but I'm not sure if I'm right.

Would you check it?

Thanks in advance
 

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Zermelo

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Jan 7, 2021
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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

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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

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Dec 15, 2005
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\(\displaystyle \frac{dy}{dx} = -x(y+1)\)

\(\displaystyle \frac{dy}{y+1} = -x \, dx\)

now integrate …
 

Timson28

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Sep 6, 2020
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omg i see my foult. Thx
 

Subhotosh Khan

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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.​
 

Jomo

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Dec 30, 2014
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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
 
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