differential equation using separable method

[sozener1]

xy'=5y

I tried to solve for general solution for the above equation

I get somthing like y=5x+c

but the answer says its y=Ax^5

why is it??

 

[HallsofIvy]

It would help if you told us how you got that answer!

(Perhaps you mis-interpreted the equation? Obviously if y= 5x+ c then xy'= 5x, not 5y.)

 

[srmichael]

xy'=5y

I tried to solve for general solution for the above equation

I get somthing like y=5x+c

but the answer says its y=Ax^5

why is it??

Without seeing your work, you most likely either separated the variables incorrectly, took the integral incorrectly or possibly did both.

\(\displaystyle xy'=5y\)

\(\displaystyle x\dfrac{dy}{dx}=5y\)

Divide both sides by x and both sides by y

\(\displaystyle \dfrac{1}{y}dy=\dfrac{5}{x}dx\)

Now proceed with the integration.

 

[sozener1]

Without seeing your work, you most likely either separated the variables incorrectly, took the integral incorrectly or possibly did both.

\(\displaystyle xy'=5y\)

\(\displaystyle x\dfrac{dy}{dx}=5y\)

Divide both sides by x and both sides by y

\(\displaystyle \dfrac{1}{y}dy=\dfrac{5}{x}dx\)

Now proceed with the integration.

I got up to where you've shown to me myself

from there I took integrals of both sides

which gave me lny=5lnx +c

from here I took exponentials of boths sides to elliminate ln

that gave me y=5x*e^c

which is different from answer

how do you get x^5 like the answer

 

[srmichael]

I got up to where you've shown to me myself

from there I took integrals of both sides

which gave me lny=5lnx +c

from here I took exponentials of boths sides to elliminate ln

that gave me y=5x*e^c

which is different from answer

Remember this exponent rule: \(\displaystyle log_{b}a^m=(m)log_{b}a\)

Apply this to 5lnx before you take the e of both sides and see what transpires.