Chapter 5: Q24E (page 166)

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

Expert verified

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

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;

$y'' = 2 x (y')^{2} y'' = \frac{d v}{d x} \frac{d v}{d x} = 2 x v^{2} \int \frac{1}{v^{2}} d v = 2 x v^{2} \int \frac{1}{v^{2}} d v = 2 \int x d x \int v^{- 2} d v = 2 \int x d x - v^{- 1} = 2 \times \frac{1}{2} x^{2} + {c_{1}}^{2} - \frac{1}{v} = x^{2} + {c_{1}}^{2} v = - \frac{1}{x^{2} + {c_{1}}^{2}} \frac{d y}{d x} = - \frac{1}{x^{2} - {c_{1}}^{2}}$

Partial Differentiation will be applied;

Using Partial fraction decomposition, we will expand the answer:

$- \frac{1}{x^{2} - {c_{1}}^{2}} = - \frac{1}{(x - c_{1}) (x + c_{1})} = \frac{(A + B) x + (A c_{1} - B c_{1})}{(x - c_{1}) (x + c_{1})} A + B = 0 A c_{1} - B c_{1} = - 1 A = - \frac{1}{2 c_{1}}, B = \frac{1}{2 c_{1}} \frac{d y}{d x} = - \frac{1}{2 c_{1}} \frac{1}{x - c_{1}} + \frac{1}{2 c_{1}} \frac{1}{x + c_{1}} \int d y = \int (- \frac{1}{2 c_{1}} \frac{1}{x - c_{1}} + \frac{1}{2 c_{1}} \frac{1}{x + c_{1}}) d x y = \frac{1}{2 c_{1}} \ln \frac{x + c_{1}}{x - c_{1}} + c_{2}$