KC Sinha Mathematics Solution Class 12 Chapter 12 द्वितीय कोटि का अवकलज (Second Order Derivative) Exercise 12.1 (Q16-Q18)

Exercise 12.1

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Question 16

यदि(If) $y=e^{-x} \cos x$ , दिखाएँ कि (show that) $\frac{d^{4} y}{d x^{4}}+4 y=0$
Sol :
$y=e^{-x} \cos x$

Differentiating w.r.t x

$\frac{d{y}}{d{x}}=e^{-x} \times(-1) \cdot \cos x+e^{-x} \times(-\sin x)$

$\frac{d y}{d x}=-e^{-x} \cos x-e^{-x} \sin x$

$\frac{d y}{d x}=-e^{-x}[\cos x+\sin x]$

Again , Differentiating w.r.t x

$\frac{d^2 y}{d x^{2}}=-e^{-x}(-1)(\cos x+\sin x)+\left(-e^{-x}\right)(-\sin x+\cos x)$

$\frac{d^{2} y}{d x^{2}}=e^{-x}[\cos x+\sin x+\sin x-\cos x]$

$\frac{d^{2} y}{d x^{2}}=2 e^{-x} \sin x$

Again , Differentiating w.r.t x

$\frac{d^{3} y}{d x^{3}}=2\left[e^{-x} \cdot(-1) \sin x+e^{-x} \cos x\right]$

$\frac{d^{3} y}{d x^{3}}=2 e^{-x}(-\sin x+\cos x)$

Again , Differentiating w.r.t x

$\frac{d^{4} y}{d x^{4}}=2\left[e^{-x} \times(-1)(-\sin x+\cos x)+e^{-x}(-\cos x-\sin x)\right]$

$\frac{d^{4} y}{d x^{4}}=2 e^{-x}[\sin x-\cos x-\cos x-\sin x]$

$\frac{d^{4} y}{d x^{4}}=2 \cdot e^{-x} \times(-2 \cos x)$

$\frac{d^{4} y}{d x^{4}}=-4 e^{-x} \cos x$

$\frac{d^{4} y}{d x^{4}}=-4 y$

$\frac{d^{4} y}{d x^{4}}+4 y=0$

Question 17

यदि (If) $e^{y}(x+1)=1$ , दिखाएँ कि (show that) $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$
Sol :
$e^{y}(x+1)=1$

Differentiating w.r.t x

$e^{y} \cdot \frac{d y}{dx}(x+1)+e^{y} \cdot 1=0$

$e^{y}\left[(x+1) \cdot \frac{d y}{d x}+1\right]=0$

$(x+1) \frac{d y}{dx}+1=0$

$\left[ \frac{d y}{d x}=\frac{-1}{x+1} \right]$

Again , Differentiating w.r.t x

$1 \cdot \frac{d y}{d x}+(x+1) \cdot \frac{d^{2} y}{d x^{2}}+0=0$

$(x+1) \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}$

$\frac{d^{2} y}{d x^{2}}=-\frac{1}{(x+1)} \cdot \frac{d y}{d x}$

$\frac{d^{2} y}{dx^{2}}=\frac{d y}{dx} \cdot \frac{dy}{dx}$

$\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$

Question 18

यदि(If) $y=e^{a \cos ^{-1} x}$ , -1≤x≤1 दिखाएँ कि (show that) $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$
Sol :
$y=e^{a \cos ^{-1} x}$

Differentiating w.r.t x

$\frac{d y}{d x}=e^{a \cos ^{-1} x} \times a \cdot \frac{-1}{\sqrt{1-x^{2}}}$

$\frac{d y}{d x}=\frac{-a e^{a \cos^{-1}}x}{\sqrt{1-x^{2}}}$

Again , Differentiating w.r.t x

$\frac{d^{2} y}{d x^{2}}=\frac{-a \cdot e^{a \cos ^{-1} x}.a \cdot \frac{1}{\sqrt{1-x^{2}}}\times \sqrt{1-x^{2}}-\left.(-a e^{\left.a \cos ^{-1} x\right.}) \times \frac{1}{2 \sqrt{1-x^{2}}}\right.\times -2x}{(\sqrt{1-x^{2}})^{2}}$

$\frac{d^{2} y}{d x^{2}}=\frac{a^{2} e^{a \cos^{-1}}x-\frac{a x \cdot e^{a \cos^{-1}}x}{\sqrt{1-x^{2}}}}{1-x^{2}}$

$\left(1-x^{2}\right) \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y+x \cdot \frac{d y}{d x}$

$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$