Question

In Problems 7–10 the independent variable x is missing in the given differential equation. Proceed as in Example 2 and solve the equation by using the substitution u = y' .

9.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{(}\mathbf{y}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$

Short answer

The solution is $\frac{1}{3}{y}^3-cy=x+k$

Step-by-step solution

Analysing the question;

We have the differential equation

$y\text{'}\text{'}+2y(y\text{'}{)}^3=0$ …(1)

Using the substitution:

First we have the substitution,

$y\text{'}=u$ …(2)

After since we have y’=u then by differentiation with respect to x, we can have

$$\begin{gathered} y\text{'}\text{'}=\frac{du}{dx} \\ =\frac{du}{dy}.\frac{dy}{dx} \\ =y\text{'}\frac{du}{dy} \\ =u\frac{du}{dy}.....\left(3\right) \end{gathered}$$

Second, substitution from equation (2) and (3) with (y’) and (y’’) into the given equation (1) then obtain

$$\begin{gathered} u\frac{du}{dy}+2y(u{)}^3=0 \\ \frac{du}{dy}+2y(u{)}^2=0 \\ \frac{du}{dy}=-2y(u{)}^2 \\ \frac{du}{u^2}=-2ydy \end{gathered}$$

Integrating both side;

$$\begin{gathered} \int \frac{du}{u^2}=\int -2ydy \\ -\frac{1}{u}=-2\times \frac{1}{2}{y}^2+c \\ -\frac{1}{u}=-y^2+c \\ {y}^2-c=\frac{1}{u} \\ u=\frac{1}{y^2-c} \end{gathered}$$

Third, substitution from equation (2) with u into equation (4), then we have;

$$\begin{gathered} y\text{'}=\frac{1}{y^2-c} \\ \frac{dy}{dx}=\frac{1}{y^2-c} \\ (y^2-c)dy=dx \end{gathered}$$

After that, we have to do integration as

$$\begin{gathered} \int (y^2-c)dy=\int dx \\ \int y^{2}dy-c\int dy=\int dx \\ \frac{1}{3}{y}^3-cy=x+k \end{gathered}$$

Hence the final solution is $\frac{1}{3}{y}^3-cy=x+k$