# If the solve the problem

Question:

If $y=e^{2 x}(a x+b)$, show that $y_{2}-4 y_{1}+4 y=0 .$

Solution:

Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$

Given,

$y=e^{2 x}(a x+b) \ldots \ldots$ equation 1

to prove: $y_{2}-4 y_{1}+4 y=0$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

As, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $d y / d x$

$\because y=e^{2 x}(a x+b)$

Using product rule to find $d y / d x$ :

$\frac{d y}{d x}=e^{2 x} \frac{d y}{d x}(a x+b)+(a x+b) \frac{d}{d x} e^{2 x}$

$\frac{d y}{d x}=a e^{2 x}+2(a x+b) e^{2 x}$

$\frac{d y}{d x}=e^{2 x}(a+2 a x+2 b) \ldots \ldots .$ equation 2

Again differentiating w.r.t $x$ using product rule:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{e}^{2 \mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{a}+2 \mathrm{ax}+2 \mathrm{~b})+(\mathrm{a}+2 \mathrm{ax}+2 \mathrm{~b}) \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{e}^{2 \mathrm{x}}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=2 \mathrm{ae}^{2 \mathrm{x}}+2(\mathrm{a}+2 \mathrm{ax}+2 \mathrm{~b}) \mathrm{e}^{2 \mathrm{x}} \ldots \ldots . .$ equation 3

In order to prove the expression try to get the required form:

Subtracting $4^{*}$ equation 2 from equation 3 :

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-4 \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{a} \mathrm{e}^{2 \mathrm{x}}+2(\mathrm{a}+2 \mathrm{ax}+2 \mathrm{~b}) \mathrm{e}^{2 \mathrm{x}}-4 \mathrm{e}^{2 \mathrm{x}}(\mathrm{a}+2 \mathrm{ax}+2 \mathrm{~b})$

$\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=2 a e^{2 x}-2 e^{2 x}(a+2 a x+2 b)$

$\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=-4 e^{2 x}(a x+b)$

Using equation 1:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-4 \frac{\mathrm{dy}}{\mathrm{dx}}=-4 \mathrm{y}$

$\therefore y_{2}-4 y_{1}+4 y=0 \ldots \ldots \ldots$ proved