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How to know if there's an explicit solution for differential equation

find_the_fun

Active member
Feb 1, 2012
166
Solve the given differential equation by separation of variables

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

I got it down to

\(\displaystyle \ln{x}x^3-\frac{x^3}{3}=y^3+3y+y^2\)

At this point I had no idea how to solve having y^3 y^2 and y terms so I did what any good student would do and checked the back of the book. The answer given was basically the same as I had got. My question is how do you know when you are done?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I get a different result:

\(\displaystyle \frac{y^2}{2}+2y+\ln|y|=\frac{x^3}{9}\left(3\ln(x)-1 \right)+C\)

I would look at the fact that we can solve for neither variable in terms of elementary functions because both are both inside and outside of log functions. So, I would leave the solution implicitly defined.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Solve the given differential equation by separation of variables

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

I got it down to

\(\displaystyle \ln{x}x^3-\frac{x^3}{3}=y^3+3y+y^2\)

At this point I had no idea how to solve having y^3 y^2 and y terms so I did what any good student would do and checked the back of the book. The answer given was basically the same as I had got. My question is how do you know when you are done?
It's a separable equation...

[tex]\displaystyle \begin{align*} y\ln{(x)}\,\frac{dx}{dy} &= \left( \frac{y + 1}{x} \right) ^2 \\ y\ln{(x)}\,\frac{dx}{dy} &= \frac{y^2 + 2y + 1}{x^2} \\ x^2\ln{(x)}\,\frac{dx}{dy} &= \frac{y^2 + 2y + 1}{y} \\ x^2\ln{(x)}\,\frac{dx}{dy} &= y + 2 + \frac{1}{y} \end{align*}[/tex]

Now you can integrate both sides with respect to y.