How to solve this ODE: $x^3dx+(y+2)^2dy=0$?

Question

I am trying to solve $$ x^3dx+(y+2)^2dy=0 quad( 1)$$

Dividing by $dx$, we can reduce the ODE to seperate variable form, i.e
$$  (1) to (y+2)^2y'=-x^3 $$
Hence,
$$ int (y(x)+2)^2y'(x) dy = int -x^3dx = - frac{x^4}{4} + c_1$$
This LHE seems to be easy to solve using integration by parts:
$$ int (y(x)+2)^2y'(x) dy = y(x)(y(x)+2)^2 - int y^3dy- int 4y^2dy + int4ydy iff$$
$$ iff - frac{y^4}{4} + -frac13 y^3 + 4y^2 + 8y + c_2$$
But then solving the ODE for $y(x)$ is a struggle:
$$ iff - frac{y^4}{4} + -frac13 y^3 + 4y^2 + 8y = - frac{3}{4}x + C$$

Any ideas on how I can solve this?

enzotib · Answer

The problem of the OP seems to be the integration by substitution.
When you set $u = y(x)$, then it follows $du = y'(x)dy$, so $y'(x)$ disappears together with $dy$ when you make the substitution and write $du$.

Ivo Terek · Answer

The coefficient of ${rm d}x$ depends only on $x$ and the coefficient of $y$ depends only on ${rm d}y$. So $$0 = x^3,{rm d}x + (y+2)^2,{rm d}y = {rm d}left(frac{x^4}{4} + frac{(y+2)^3}{3}right),$$which says that your solutions are described by $$frac{x^4}{4} + frac{(y+2)^3}{3} = c,$$for some $c in Bbb R$. Thus $$y = -2+left(frac{c-3x^4}{4}right)^{1/3},$$after letting $c$ absorb some constants.

Answered by Ivo Terek on November 20, 2021

Sameer Baheti · Answer

Substitute $Y=y+2$ initially. Solving $x^3dx+Y^2dY=0$ yields $Y=left(C^prime-frac{3x^4}4right)^{frac13}=y+2$.

user2715281 · Answer

$$x^3dx+(y+2)^2dy=0$$
$$x^3dx=-(y+2)^2dy$$
$$x^3=-(y+2)^2y'$$
Integratation gives:
$$int x^3dx=-int (y+2)^2 y'dx$$
$$int x^3dx=-int (y+2)^2 dy$$
Substitute $u=y+2$
$$int x^3dx=-int u^2du$$
$$dfrac {x^4}4+dfrac {u^3}{3}=C$$
$$dfrac {x^4}4+dfrac {(y+2)^3}{3}=C$$

How to solve this ODE: $x^3dx+(y+2)^2dy=0$?

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