Question

Find the general solution of the following function by setting y/x = u

Answer 1

\[x y' = y^2 + y\]

Answer 2

\[\large xy'=y^2+y\]Dividing both sides by x gives us,
\[\large y'=\frac{y}{x}y+\frac{y}{x}\]

Applying our substitution gives us,\[\large y'=uy+u\]Do you think you can solve it from here? :)
It appears to be a separable equation.

Answer 3

Should I replace y' with (u'x + u) and y with xu ?

Answer 4

Mmmmmm no probably not.
I would leave it like this, as an equation in y and u.

Separate the y's to one side, and the u's to the other.
And then integrate with respect to each variable.

Answer 5

Understand how to separate them?\[\large y'=uy+u \qquad \rightarrow \qquad y'=u(y+1) \qquad \rightarrow \qquad \frac{1}{y+1}y'=u\]

Answer 6

ah okay I see now. now this--> ?
\[\int\limits_{ }^{ } \frac{1}{y+1}\text{ dy} = \int\limits_{ }^{ }u \text{ du}\]

Answer 7

yes

Answer 8

and that evaluates to
\[\ln {(y+1)} = u^2/2 +c_1\\y+1 = e^{u^2+c_1}\\y=c_2e^{u^2}-1\\y=c_2e^{(y/x)^2}-1\]?

Answer 9

Yah that looks right :) Good job!

Answer 10

Thanks for the help @zepdrix !