Solving Differential Equations: Integrating (y+1)^2/y dy = x^2 ln x dx

[basty]

I have differential equations problem, the problem is:

$\frac{(y+1)^2}{y} dy = x^2 \ln x \ dx$

Integrating both sides will yield:

$\frac{1}{2} y^2 + 2y + \ln y = \ln x \ . \frac{1}{3}x^3 - \frac{1}{9} x^3 + c$

Is this the final solution?

If not, what is the final solution?

 

[Mark44]

I have differential equations problem, the problem is:

$\frac{(y+1)^2}{y} dy = x^2 \ln x \ dx$

Integrating both sides will yield:

$\frac{1}{2} y^2 + 2y + \ln y = \ln x \ . \frac{1}{3}x^3 - \frac{1}{9} x^3 + c$

Is this the final solution?

If not, what is the final solution?

Assuming your work is correct (I didn't check), what you got looks fine. You can check for yourself by differentiating both sides with respect to x. If what you ended with is correct, you should be able to get back to the equation at the beginning of your post.