Question

Verify that 𝑦1(𝑥) = 1 and 𝑦2(𝑥) = 𝑥^(1/2) are solutions of the differential equation 𝑦𝑦′′ + (𝑦′)^2 = 0 for 𝑥 > 0. Then show that 𝑦 = 𝑐1 + 𝑐2𝑥^(1/2) is not, in general, a solution to the equation. Explain why this does not contradict superposition principle.

Answer 1

Compute derivatives of $y_1(x)$ : $(y_1(x))'=0$, $(y_1(x))''=0.$ Thus, we get: $y_1(x)(y_1(x))''+(y_1(x)')^2=0$. Compute derivatives of $y_2(x)$: $(y_2(x))'=\frac12x^{-\frac12}$, $(y_2(x))''=-\frac14x^{-\frac32}$. We get: $y_2(x)(y_2(x))''+(y_2(x)')^2=0$.

Consider the function $y=c_1+c_2x^{\frac12}$. Compute the derivatives: $y'=\frac12c_2x^{-\frac12}$, $y''=-\frac14c_2x^{-\frac32}$. Consider the expression: $y(x)(y(x))''+(y')^2=-\frac14c_1c_2x^{-\frac32}-\frac14c_2^2x^{-1}+\frac14c_2^2x^{-1}=-\frac14c_1c_2x^{-\frac32}$

As we can see, $y(x)$ is a solution if and only if either $c_1=0$ or $c_2=0$. It means that in general $y(x)$ is not the solution. It is due to the fact that the equation $yy''+(y')^2=0$ is nonlinear. Linear superposition principle holds only for linear differential equations.