KC Sinha Mathematics Solution Class 12 Chapter 12 द्वितीय कोटि का अवकलज (Second Order Derivative) Exercise 12.1 (Q22-Q24)

Exercise 12.1

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Question 22

यदि(If) $y=x^{n-1} \log x$ सिद्ध करे कि(prove that) $x^{2} y_{2}+(3-2 n) x y_{1}+(n-1)^{2} y=0$
Sol :
$y=x^{n-1} \cdot \log x$

Differentiating w.r.t x

$y_{1}=(n-1) x^{n-2} \cdot \log x+x^{n-1} \cdot \frac{1}{x}$

$y_{1}=(n-1) x^{n-2} \cdot \log x+x^{n-2}$

Again , Differentiating w.r.t x

$y_{2}=(n-1)\left[(n-2) x^{n-3} \cdot \log x+x^{n-2} \times \frac{1}{x}\right]+(n-2) x^{n-3}$

$y_{2}=(n-1)\left[(n-2) x^{n-3} \log x+x^{n-3}\right]+(n-2) x^{n-3}$

$y_2=\left(n^{2}-3 n+2\right) \cdot x^{n-3} \cdot \log x+(n-1) x^{n-3}+(n-2) \cdot x^{n-3}$

$y_{2}=\left(n^{2}-3 n+2\right) x^{n-3} \cdot \log x+(2 n-3) \cdot x^{n-3}$

L.H.S

$x^{2} y_{2}+(3-2 n) x y_{3}+(n-1)^{2}$

$=x^{2}\left[\left(n^{2}-3 n+2\right) x^{n-3} \cdot \log x+(2 n-3) x^{n-3}\right]+(3-2 n) x[(n-1)x^{n-2}.\log x+x^{n-2}]+(n^2-2n+1)x^{n-1}\log x$

$=\left(n^{2}-3 n+2\right) x^{n-1} \log x+(2 n-3) x^{n-1}+\left(5 n-3-2 n^{2}\right) x^{n-1} \cdot \log x+(3-2 n) \cdot x^{n-1}+(n^2-2n+1)x^{n-1}\log x$

$=\left[n^{2}-3 n+2+5-3-2 n^{2}+n^{2}-2 n+1\right] x^{n-1} \log x+(2n-3+3-2n)x^{n-1}$

$=\left[2 n^{2}-2 n^{2}-5 n+5 n+3-3\right] \cdot x^{4-1} \cdot \log x+0 \cdot x^{n-1}$

=0

Question 23

यदि(If) $y=\log \left(\frac{x}{a+b x}\right)^{x}$

सिद्ध करे कि (prove that) $\frac{d^{2} y}{d x^{2}}=\frac{1}{x}\left(\frac{a}{a+b x}\right)^{2}$

या यदि (if) $x=(a+b x) e^{\frac{y}{x}}$ सिद्ध करे कि (prove that)

$x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d y}{d x}-y\right)^{2}$
Sol :
$\frac{y}{x}=\log \left(\frac{x}{a+b x}\right)$

$e^{\frac{y}{2}}=\frac{x}{a+b z}$

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

$y=\log \left(\frac{x}{a+b x}\right)^{x}$

$y=x \cdot \log \left(\frac{x}{a+b x}\right)$

y=x[log x-log (a+bx)]

$\frac{y}{x}=\log x-\log (a+b x)$

Differentiating w.r.t x

$\frac{\frac{d y}{dx} x-y \cdot 1}{x^{2}}=\frac{1}{x}-\frac{1}{a+b x} \times b$

$x \frac{d y}{dx}-y=x^{2}\left[\frac{1}{x}-\frac{b}{a+b x}\right]$

$\frac{x{dy}}{dx}-y=x^{2}\left[\frac{a+b x-b x}{x(a+b x)}\right]$

$x \frac{d y}{d x}-y=\frac{a x}{a+b x}$

Differentiating w.r.t x

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

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

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

Multiplying by $x^{2}$ in both sides

$x^{3} \cdot \frac{d^{2} y}{d x^{2}}=\frac{a^{2} x^{2}}{(a+b x)^{2}}$

$x^{3} \cdot \frac{d^{2} y}{d x^{2}}=\left(\frac{a x}{a+b x}\right)^{2}$

$x^{3} \frac{d^{2} y}{d x^{2}}=\left(x \frac{d x}{d x}-y\right)^{2}$

Question 24

यदि(If) x=cosθ , $y=\sin ^{3} \theta$ , दिखाएँ कि(show that) $y\frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=3 \sin ^{2} \theta\left(5 \cos ^{2} \theta-1\right)$
Sol :
x=cosθ , $y=\sin ^{3} \theta$

Differentiating w.r.t x

$\frac{d x}{d \theta}=-\sin \theta$..(i)

$\frac{d y}{d \theta}=3 \sin^2 \theta \cos \theta$..(ii)

समीकरण (ii) मे (i) से भाग देने पर

$\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3\sin^2 \theta \cos \theta}{-\sin \theta}$

$\frac{d y}{d x}=-3 \sin \theta \cos \theta$

Again ,Differentiating w.r.t x

$\left.\frac{d^{2} y}{dx^{2}}=-3 \Big[\cos \theta \cdot \frac{d \theta}{d x} \cdot \cos \theta+\sin \theta(-\sin \theta) \frac{d \theta}{d x}\right]$

$\frac{d^{2} y}{d x^{2}}=-3\left[\cos ^{2} \theta-\sin ^{2} \theta\right] \frac{d \theta}{d x}$

$\frac{d^{2} y}{d x^{2}}=-3\left[\cos^{2} \theta-\sin^{2} \theta\right] \frac{1}{-\sin \theta}$

$\frac{d^{2} y}{d x^{2}}=\frac{3}{\sin \theta}\left(\cos^{2} \theta-\sin^2 \theta\right)$

L.H.S

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

$=\sin ^{3} \theta \frac{3}{\sin \theta}\left(\cos ^{2} \theta-\operatorname{sin}^{2} \theta\right)+(-3 \sin \theta \cos \theta)^{2}$

$=3 \sin ^{2} \theta-\cos ^{2} \theta-3 \sin ^{2} \theta \cdot \sin ^{2} \theta+9 \sin ^{2} \theta \cos ^{2} \theta$

$=3 \sin ^{2} \theta \cos ^{2} \theta-3 \sin ^{2} \theta\left(1-\cos ^{2} \theta\right) + 9 \sin ^{2} \theta \cos^2 \theta$

$=3 \sin ^{2} \theta \cos ^{2} \theta-3 \sin ^{2} \theta+3 \sin ^{2} \theta \cos ^{2} \theta+9\sin^2 \theta \cos ^2 \theta$

$=15 \mathrm{sin}^{2} \theta \cos^2 \theta-3\sin^2 \theta$

$=3 \sin ^{2} \theta\left[5\cos^{2} \theta-1\right]$