Question

Show that the substitution u = y’ leads to a Bernoulli equation. Solve this equation

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

Short answer

The solution is$y=-\ln \left|c_1-x^2\right|+c_2$

Step-by-step solution

Substituting these values

Let’s substitute u=y’ . Then, y’’=u’. Substitute these values into given equation

$$\begin{gathered} xy\text{'}\text{'}=y+x{(y\text{'})}^{2}xu\text{'}=u+x{u}^2 \\ u\text{'}=\frac{u}{x}+u^2 \\ u\text{'}-\frac{1}{x}u-u^2 \end{gathered}$$

Which is a Bernoulli’s equation with valuse of n=2.

Substitute to reduce equation to a linear one;

We substitute $\omega =u^{1-2}=u^{-1}$or $u={\omega }^{-1}$to reduce the equation into linear;

$$\begin{gathered} \frac{du}{dx}=\frac{du}{d\omega }\frac{d\omega }{dx}=-{\omega }^2\frac{d\omega }{dx} \\ -{\omega }^2\frac{d\omega }{dx}+\frac{1}{x}{\omega }^{-1}={\left({\omega }^{-1}\right)}^2{\omega }^{-2}\frac{d\omega }{dx}+\frac{1}{x}{\omega }^{-1}={\omega }^{-2} \\ \frac{d\omega }{dx}+\frac{1}{x}\omega =-1 \end{gathered}$$

Multiplying the equation to get integrating factor;

multiply the equation with the integrating factor to get

$x\frac{d\omega }{dx}+\omega =-x$

Integrating we get

$$\begin{gathered} \int \frac{d}{dx}\left[x\omega \right]=\int -xdx \\ x\omega =-\frac{1}{2}{x}^2+c \\ \omega =-\frac{x}{2}+\frac{c}{x} \end{gathered}$$

since, $\omega =u^{-1}=\frac{1}{u}$and $u=y\text{'}$the solution of the given equation will be

Hence the final answer is$y=-\ln \left|c_1-x^2\right|+c_2$