How can i found the solution for this: solve this using Lagrange method (differential equation): $$\color{red}{y=2y'x+y^3}$$ Hint: $$\color{#00F}{y^3 \equiv y'''}$$

Can you explain the method?

This is a capture for this example with solution. enter image description here


Find the general solution for this differential equation


This equation is a differential Lagrange equation, by integrate the 02 (two) parts we obtain:


  • $\begingroup$ The equation in red is separable. $\endgroup$ – Bitrex Apr 1 '14 at 20:10
  • $\begingroup$ In your differential equation, does $y^3$ denote $\frac{d^3y}{dy^3}$ or $y\cdot y\cdot y$? $\endgroup$ – Jason Zimba Apr 1 '14 at 20:13
  • $\begingroup$ $y^3$ denote $\frac{d^3y}{dy^3}$ $\endgroup$ – Safe Mode Apr 1 '14 at 20:32
  • $\begingroup$ Where is the solution? $\endgroup$ – Safe Mode Apr 1 '14 at 22:37
  • $\begingroup$ @Bitrex OP says that $y^3$ is the third derivative. $\endgroup$ – Jason Zimba Apr 2 '14 at 1:39

We are given (I think there is an issue in the hint):

$$\tag 1 y = 2y' x + y^3, y^3 \equiv (y')^3$$

If we let $p = \dfrac{dy}{dx}$, we can rewrite $(1)$ as:

$$\tag 2 y = 2px + p^3$$

Differentiating $(2)$ wrt $x$ yields:

$$\dfrac{dy}{dx} = p = 2p + 2 x \dfrac{dp}{dx} + 3 p^2 \dfrac{dp}{dx}$$

This reduces to:

$$\tag 3 3p^2\frac{dp}{dx} + 2x\frac{dp}{dx} + p = 0$$

Solving $(3)$ yields four solutions as:

$$p(x) = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$

From our initial substitution, we have $p = \dfrac{dy}{dx}$, so we can write:

$$\dfrac{dy}{dx} = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$

We can solve these four (it is possible that not all four satisfy the ODE) for $y(x)$ by rearranging and integrating. For example:

$$\int dy = -\sqrt{\frac{2}{3}} \int \sqrt{~-\sqrt{c_1+x^2}-x}~~dx $$

where $c_1$ is a constant.

  • $\begingroup$ I can understand that! $\endgroup$ – Namaste Apr 3 '14 at 12:24

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