Question

Verify that y² = 4ax is a solution of the differential equation y = x $\frac{dy}{dx}+a\frac{dx}{dy}$.

Answer 1

We have,
$y^2=4ax$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$2y\frac{dy}{dx}=4a$
$\Rightarrow$$\frac{dy}{dx}=\frac{2a}{y}$ ...(2)
Now, differentiating both sides of (1) with respect to $y$, we get
$2y=4a\frac{dx}{dy}$
$\Rightarrow$$\frac{dx}{dy}=\frac{y}{2a}$ ...(3)
$$\begin{gathered} \therefore x\frac{dy}{dx}+a\frac{dx}{dy}=x\left(\frac{2a}{y}\right)+a\left(\frac{y}{2a}\right)\left[\text{Using}\left(2\right)\text{and}\left(3\right)\right] \\ \Rightarrow x\frac{dy}{dx}+a\frac{dx}{dy}=\frac{2ax}{y}+\frac{y}{2} \\ \Rightarrow x\frac{dy}{dx}+a\frac{dx}{dy}=\frac{y^2}{2y}+\frac{y}{2}\left[\text{Using}\left(1\right)\right] \\ \Rightarrow x\frac{dy}{dx}+a\frac{dx}{dy}=\frac{y}{2}+\frac{y}{2} \\ \Rightarrow x\frac{dy}{dx}+a\frac{dx}{dy}=y \\ \text{⇒}y=x\frac{dy}{dx}+a\frac{dx}{dy} \end{gathered}$$
Hence, the given function is the solution to the given differential equation.