Question

Discuss how to find an alternative two-parameter family of solutions for the nonlinear differential equation y’’ = 2x( y’)2 in Example 1.

Short answer

The solution of the above question is $y=\frac{1}{2c_1}\ln \frac{x+c_1}{x-c_1}+c_2$.

Step-by-step solution

Solving of second order and non-linear differential equation:

Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions. To decompose a fraction, you first factor the denominator.

Method of reduction order will be applied:

In this question we will apply the concept Method of reduction order;

$$\begin{gathered} y\text{'}\text{'}=2x(y\text{'}{)}^2 \\ y\text{'}\text{'}=\frac{dv}{dx} \\ \frac{dv}{dx}=2x{v}^2 \\ \int \frac{1}{v^2}dv=2x{v}^2 \\ \int \frac{1}{v^2}dv=2\int xdx \\ \int v^{-2}dv=2\int xdx \\ -v^{-1}=2\times \frac{1}{2}{x}^2+{c_1}^2 \\ -\frac{1}{v}=x^2+{c_1}^2 \\ v=-\frac{1}{x^2+{c_1}^2} \\ \frac{dy}{dx}=-\frac{1}{x^2-{c_1}^2} \end{gathered}$$

Partial Differentiation will be applied;

Using Partial fraction decomposition, we will expand the answer:

$$\begin{gathered} -\frac{1}{x^2-{c_1}^2}=-\frac{1}{\left(x-c_1\right)\left(x+c_1\right)} \\ =\frac{(A+B)x+(A{c}_1-B{c}_1)}{\left(x-c_1\right)\left(x+c_1\right)} \\ A+B=0 \\ A{c}_1-B{c}_1=-1 \\ A=-\frac{1}{2c_1},B=\frac{1}{2c_1} \\ \frac{dy}{dx}=-\frac{1}{2c_1}\frac{1}{x-c_1}+\frac{1}{2c_1}\frac{1}{x+c_1} \\ \int dy=\int \left(-\frac{1}{2c_1}\frac{1}{x-c_1}+\frac{1}{2c_1}\frac{1}{x+c_1}\right)dx \\ y=\frac{1}{2c_1}\ln \frac{x+c_1}{x-c_1}+c_2 \end{gathered}$$

Hence, the final answer is $y=\frac{1}{2c_1}\ln \frac{x+c_1}{x-c_1}+c_2$.