Continuing after integrating in a differential equation problem

Question

I have this problem below:

$dy/dx=x^2/y^2$

With what I have

$1/x^2=1/y^2$

$ \int\ 1/x^2 dx= \int\ 1/y^2 dy$

how do you continue?

Answer 1

$\int\ 1/x^2 dx= \int\ 1/y^2 dy$ is wrong. From $dy/dx=x^2/y^2$ you get $\int\ x^2 dx= \int\ y^2 dy$, hence $y^3=x^3+C$.

Can you proceed ?

Answer 2

Your book is right $$\frac {dy}{dx} = h(x)g(y)$$ $$\int \frac {dy}{g(y)} = \int h(x)dx$$ But for $$ \frac {dy}{dx} =\frac { h(x)}{g(y)}$$ We have $$ \int {g(y)} dy =\int { h(x)}{dx}$$ You can also apply the rule of the book on this
$$ \frac {dy}{dx} =\frac { h(x)}{s(y)}$$ Apply the rule's book with $g(y)=\frac 1 {s(y)}$ $$\frac {dy}{dx} = h(x)g(y) \to \int \frac {dy}{g(y)} = \int h(x)dx$$ $$ \frac {dy}{dx} =\frac { h(x)}{s(y)} \to \int \frac {dy}{\frac 1 {s(y)}} = \int h(x)dx $$ $$\int \frac {dy}{\frac 1 {s(y)}} = \int h(x)dx \to \int {s(y)} dy = \int h(x)dx $$ It gives the same answer ...