u′ = 1 + u2 separation of variable
Answers
This is already in the required form (since the x-terms are together with dx terms, and y-terms are together with dy terms), so we simply integrate:
\displaystyle\int{y}^{2}{\left.{d}{y}\right.}+\int{x}^{3}{\left.{d}{x}\right.}={0}∫y
2
dy+∫x
3
dx=0
Giving:
\displaystyle\frac{{y}^{3}}{{3}}+\frac{{x}^{4}}{{4}}={K}
3
y
3
+
4
x
4
=K
This is the general solution for the differential equation.
We can continue on to solve this as an explicit function in x, as follows:
\displaystyle{y}^{3}={3}{\left({K}-\frac{{x}^{4}}{{4}}\right)}y
3
=3(K−
4
x
4
)
\displaystyle{y}={\sqrt[{{3}}]{{{3}{\left({K}-\frac{{x}^{4}}{{4}}\right)}}}}y=
3
3(K−
4
x
4
)
Taking a typical constant value \displaystyle{K}={5}K=5, we have this solution graph: