Question

if 2y=x(1+dy/dx) show that d2y/ dx2 is constant​

Answer 1

Given,

$\longrightarrow2y=x\left(1+\dfrac{dy}{dx}\right)$

$\longrightarrow\dfrac{2y}{x}=1+\dfrac{dy}{dx}$

$\longrightarrow\dfrac{dy}{dx}=\dfrac{2y}{x}-1\quad\quad\dots(1)$

Differentiating,

$\longrightarrow\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)=\dfrac{d}{dx}\left(\dfrac{2y}{x}-1\right)$

$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{2x\cdot\dfrac{dy}{dx}-2y}{x^2}$

From (1),

$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{2x\left(\dfrac{2y}{x}-1\right)-2y}{x^2}$

$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{4y-2x-2y}{x^2}$

$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{2(y-x)}{x^2}\quad\quad\dots(2)$

But we can rewrite (1) as,

$\longrightarrow\dfrac{dy}{dx}-\dfrac{2y}{x}=-1$

We are having a first order linear differential equation. So,

• $P(x)=-\dfrac{2}{x}$

• $Q(x)=-1$

Then integrating factor,

$\longrightarrow IF=e^{\int P(x)\ dx}$

$\longrightarrow IF=e^{-2\int\frac{1}{x}\ dx}$

$\longrightarrow IF=e^{-2\log|x|}$

$\longrightarrow IF=\dfrac{1}{x^2}$

Then, the solution will be,

$\displaystyle\longrightarrow y\cdot IF=\int Q(x)\cdot IF\ dx$

$\displaystyle\longrightarrow\dfrac{y}{x^2}=-\int\dfrac{1}{x^2}\ dx$

$\displaystyle\longrightarrow\dfrac{y}{x^2}=\dfrac{1}{x}+C$

$\displaystyle\longrightarrow\dfrac{y}{x^2}-\dfrac{1}{x}=C$

$\displaystyle\longrightarrow\dfrac{y-x}{x^2}=C$

Then (2) becomes,

$\longrightarrow\underline{\underline{\dfrac{d^2y}{dx^2}=2C}}$

Hence Proved that $\dfrac{d^2y}{dx^2}$ is a constant.