Question

Show that the function $y=Ax+\dfrac{B}{x}$ is a solution of the differential equation ${{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}-y=0$.

Answer 1

For solving this question you should know about the differential equations and to find the second order derivative of any function. In this question we will find the first order derivative of y and then find the values of A and B and then put these after finding the second order derivative in it. And then we will solve it.

Complete step-by-step solution:
So, according to our question it is asked of us to prove that a function $y=Ax+\dfrac{B}{x}$ is a solution of the differential equation ${{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}-y=0$. So, if it is given then it will be necessary that the function will come if we solve this equation and it is also possible that if we solve the equation then you will get the differential equation. So, if we see that the given function is,
$y=Ax+\dfrac{B}{x}\ldots \ldots \ldots \left( 1 \right)$
Now, if we differentiate it on both sides, then,
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=A+B\left( \dfrac{-1}{{{x}^{2}}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=A-\dfrac{B}{{{x}^{2}}}\ldots \ldots \ldots \left( 2 \right) \\\ \end{aligned}$
Now, the equation is in the second order, so we will differentiate it to the first order derivative again. So, we will get,
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2B}{{{x}^{3}}} \\\ & \Rightarrow B=\dfrac{{{x}^{3}}}{2}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\ldots \ldots \ldots \left( 3 \right) \\\ \end{aligned}$
Now if we substitute this is equation (2), then we get,
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=A-\dfrac{x}{2}\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \\\ & \Rightarrow A=\dfrac{dy}{dx}+\dfrac{x}{2}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\ldots \ldots \ldots \left( 4 \right) \\\ \end{aligned}$
Now, we will put this value in equation (1) again to get the value of y. So, by substituting (4) and (3) in equation (1), we get,
$\begin{aligned} & \Rightarrow y=\left( \dfrac{dy}{dx}+\dfrac{x}{2}\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)x+\left( \dfrac{{{x}^{3}}}{2}\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\dfrac{1}{x} \\\ & \Rightarrow {{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}-y=0 \\\ \end{aligned}$
So, it is determined as the same as given in the equation. So, it is proved that $y=Ax+\dfrac{B}{x}$ is a solution of ${{x}^{2}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}-y=0$.

Note: For solving these types of questions we have to mind that if any equation is satisfying to each other then it is confirmed that they both will be solved by each other or they will be solutions of each other and thus we will prove that.