#### Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 23

Proved

Hint:

You must know about the derivative of exponential function

Given:

$y=e^{2x}(ax+b), \: \: show\: \: that\: \: y_{2}-4y_{1}+4y=0$

Solution:

$y=e^{2x}(ax+b)\; \; \; \; \; \; \; \; \; ......(1)$

\begin{aligned} &\text { By using product rule of derivation }\\ &\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) \end{aligned}$.....(2)$

\begin{aligned} &\text { Again differentiating both sides w.r.t } x, \text { using product rule }\\ &\frac{d^{2} y}{d x^{2}}=e^{2 x} \frac{d y}{d x}(a+2 a x+2 b)+(a+2 a x+2 b) \frac{d}{d x} e^{2 x}\\ &\frac{d^{2} y}{d x^{2}}=2 e^{2 x}+2(a+2 a x+2 b) e^{2 x} \quad \ldots .(3) \end{aligned}

$In\; order\; to\; prove\; the\; expression\; try\; to\; get\; the\; required\; f\! orm \\Subtracting\; 4 \times equation\; (2)\; f\! rom\; equation (3)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=2 e^{2 x}+2(a+2 a x+2 b) e^{2 x}-4 e^{2 x}(a+2 a x+2 b)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=2 a e^{2 x}-2 e^{2 x}(a+2 a x+2 b)\\ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=-4 e^{2 x}(a x+b)$

\begin{aligned} &\text { Using equation }(1)\\ &\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}=-4 y\\ &\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=0\\ &y_{2}-4 y_{1}+4 y=0 \end{aligned}