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[Find the second order derivatives of the following functions]
(i) log x
Sol :
माना y=log x
Differentiating w.r.t x
$\frac{d y}{dx}=\frac{1}{x}$
Again, Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-\frac{1}{x^{2}}$
(ii) x20
Sol :
माना y=x20
Differentiating w.r.t x
$\frac{d y}{dx}=20 x^{19}$
Again , differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=380 x^{18}$
(iii) log(log x)
Sol :
माना y=log(log x)
Differentiating w.r.t x
$\frac{d y}{d x}=\frac{1}{\log x} \times \frac{1}{x}$
$\frac{d y}{d x}=\frac{1}{x \log x}$
Again , Differentiating w.r.t x
$\frac{d^{2}y}{d x^{2}}=\frac{0 \cdot x \log x-1 \cdot\left[1 \cdot \log {x}+x \times \frac{1}{x}\right]}{(x \log x)^{2}}$
$=\frac{-\left(\log x+1\right)}{\left(x \log x\right)^{2}}$
(iv) x2+3x+2
Sol :
माना y=x2+3x+2
Differentiating w.r.t x
$\frac{d y}{d x}=2 x+3$
Again, differentiating w.r.t x
$\frac{d^{2} y}{dx^{2}}=2$
(v) xcosx
Sol :
माना y=xcosx
Differentiating w.r.t x
$\frac{d y}{d x}=1 \cdot \cos x+x(-\sin x)$
$\frac{d y}{d x}=\cos x-x \sin x$
Again, Differentiate w.r.t x
$\frac{d^{2} y}{d x^{2}}=-\sin x-[1 \cdot \operatorname{sin} x+x \cos x]$
=-sinx-sinx-xcosx
=-xcosx-2sinx
(vi) exsin5x
Sol :
माना y=exsin5x
Differentiating w.r.t x
$\frac{d y}{dx}=e^{x} \cdot \sin 5 x+e^{x} \cdot \cos 5x \times 5$
$\frac{d y}{d x}=e^{x} \sin 5 x+5e ^{2} \cos x$
Again, Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=e^{x} \sin 5 x+e^{x} \cos 5 x \times 5+\left[e^{x} \cos 5 x+e^{2}(-\sin 5x) \times 5\right]$
=exsin5x+5excos5x+5excos5x-25exsin5x
=10excos5x-24exsin5x
=2ex(5cos5x-12sin5x)
(vii) sin(log x)
Sol :
माना y=sin(log x)
Differentiating w.r.t x
$\frac{d y}{dx}=\cos (\log x) \times \frac{1}{x}$
$\frac{d y}{dx}=\frac{\cos (\log x)}{x}$
Again, Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=\frac{-\sin (\log x) \times \frac{1}{x} \times x-\cos (\log x) \times 1}{x^{2}}$
$=-\frac{\sin\left(\log x\right)+\cos \left(\log x\right)}{x^{2}}$
Contents
Question 2
यदि(If) y=sin(log x) , सिद्ध करे कि (prove that) $x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0$
Sol :
y=sin(log x)
Differentiating w.r.t x
$\frac{d y}{d x}=\cos (\log x) \times \frac{1}{x}$
$\frac{dy}{dx}=\frac{\cos\left(\log x\right)}{x}$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=\frac{-\sin \left(\log x\right) \times \frac{1}{x}.x-\cos \left(\log x\right) \cdot 1}{x^{2}}$
$\frac{d^{2} y}{d x^{2}}=-\frac{[\sin 4(\log x)+\cos (\log x)]}{x^{2}}$
$x^{2} \frac{d^{2} y}{d x^{2}}=-\sin (\log x)-\cos (\log x)$
$x^{2} \frac{d^{2} y}{d x^{2}}=-y-x \cdot \frac{d y}{dx}$
$x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{dx}+y=0$
Question 3
यदि(If) y=x3+tanx , दिखाएँ कि(show that) $\frac{d^{2} y}{d x^{2}}=6 x+2 \sec ^{2} x \tan x$
Sol :
y=x3+tanx
Differentiating w.r.t x
$\frac{d y}{d x}=3 x^{2}+\sec ^{2} x$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=6 x+2 \sec x \cdot \sec x \tan x$
$=6 x+2 \operatorname{sec}^{2} x \tan x$
Question 4
यदि(If) y=sinx(sinx) , दिखाएँ कि(show that) $\frac{d^{2} y}{d x^{2}}+\tan x \frac{d y}{d x}+y \cos ^{2} x=0$
Sol :
y=sin(sinx)
Differentiating w.r.t x
$\frac{d y}{d x}=\cos (\sin x)\times \cos x$
$\cos(\sin x)=\frac{1}{\cos x} \cdot \frac{dy}{d x}$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-\sin (\sin 3 x) \cdot \cos x \cdot \cos x+\cos (\sin x) \times(-\sin x)$
$\frac{d^{2} y}{d x^{2}}=-y \cos ^{2} x \cdot+\frac{1}{\cos x} \cdot \frac{d y}{d x}(-\sin x)$
$\frac{d^{2} y}{d x^{2}}=-y \cos ^{2} x-\tan x \frac{d y}{dx}$
$\frac{d^{2} y}{d x^{2}}+\tan x \frac{d y}{d x}+y\cos^{2} x=0$
Question 5
यदि(If) $y=[\log (x+\sqrt{x^{2}+a^{2}})]^{2}$ , सिद्द करे कि(prove that) $\left(x^{2}+a^{2}\right) y_{2}+x y_{1}=2$
Sol :
$y=\left[\log(x+\sqrt{x^{2}+a^{2}})\right]^{2}$
Differentiating w.r.t x
$y_{1}=2\left[\log (x+\sqrt{x^{2}+a^{2}}) \cdot \frac{1}{x+\sqrt{x^{2}+a^{2}}} \times\left(1+\frac{1}{2 \sqrt{x^{2}+a^{2}}} \times 2 x\right.\right)$
$y_{1}=2 \frac{\left[\log(x+\sqrt{x^{2}+a^{2}}\right]}{x+\sqrt{x^{2}+a^{2}}}\left[\frac{\sqrt{x^{2} + a^{2}}+x}{\sqrt{x^{2}+a^{2}}}\right]$
$y_{1}=\frac{2[\log (x+\sqrt{x^{2}+a^{2}})]}{\sqrt{x^{2}+a^{2}}}$
Again ,Differentiating w.r.t x
$y_{2}=\frac{2 \cdot \frac{1}{x+\sqrt{x^{2}+a^{2}}}\left[1+\frac{1}{2 \sqrt{x^{2}+a^{2}}} \times 2 x\right] \cdot \sqrt{x^{2}+a^{2}}-2[\log(x+\sqrt{x^2+a^2})]\times \frac{1}{2\sqrt{x^2+a^2}}\times 2x }{(\sqrt{x^{2}+a^{2}}]^{2}}$
$y_{2}=\frac{\frac{2}{x+\sqrt{x^{2}+a^{2}}}\left[\frac{\sqrt{x^{2}+a^{2}}+x}{\sqrt{x^{2}+a^{2}}}\right] \cdot \sqrt{x^{2}+a^{2}}-\frac{2 \times \log (x+\sqrt{x^{2}+a^{2}}}{\sqrt{x^{2}+a^{2}}}}{x^{2}+a^{2}}$
$\left(x^{2}+a^{2}\right) y_{2}=2-xy_{1}$
$\left(x^{2}+a^{2}\right) y_{2}+x y_{1}=2$
Question 6
यदि(If) $y=\left(1-x^{2}\right)^{3 / 2}$ , दिखाएँ कि (show that) $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+3 y=0$
Sol :
$y=\left(1-x^{2}\right)^{\frac{3}{2}}$
Differentiating w.r.t x
$\frac{d y}{d x}=\frac{3}{2}\left(1-x^{2}\right)^{\frac{1}{2}} \cdot[0-2 x]$
$\frac{d y}{dx}=\frac{3}{2}\left(1-x^{2}\right)^{\frac{1}{2}} \cdot(-2 x)$
$\frac{dy}{d x}=-3 x\left(1-x^{2}\right)^{\frac{1}{2}}$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-3 \cdot\left(1-x^{2}\right)^{\frac{1}{2}}+(-3 x) \frac{1}{2}\left(1-x^{2}\right)^{-\frac{1}{2}} \times(-2 x)$
$\frac{d^{2} y}{d x^{2}}=-3\left(1-x^{2}\right)^{\frac{1}{2}}+\frac{3 x^{2}}{\left(1-x^{2}\right)^{\frac{1}{2}}}$
Multiplying by $\left(1-x^{2}\right)$ in both sides
$\left(1-x^{2}\right) \cdot \frac{d^{2} y}{d t^{2}}=-3\left(1-x^{2}\right)^{\frac{3}{2}}+3 x^{2} \cdot\left(1-2^{2}\right)^{\frac{1}{2}}$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=-3 y-x\left[-3 x\left(1-x^{2}\right)^{\frac{1}{2}}\right]$
$\left(1-x^{2}\right) \cdot \frac{d^{2} y}{d x^{2}}=-3 y-x \frac{d y}{d x}$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+3 y=0$
Question 7
यदि(If) y=Asinx+Bcosx, सिद्द करे कि (prove that) $\frac{d^{2} y}{d x^{2}}+y=0$
Sol :
y=Asinx+Bcosx
Differentiating w.r.t x
$\frac{d y}{d x}=A \cos x+B(-\sin x)$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=A(-\sin x)-B\cos x$
$\frac{d^{2} y}{d x^{2}}=-\left[A \sin x+B\cos x\right]$
$\frac{d^{2} y}{d x^{2}}=-y$
$\frac{d^{2} y}{d x^{2}}+y=0$
Question 8
यदि(If) y=5cosx-3sinx , सिद्ध करे कि (prove that) $\frac{d^{2} y}{d x^{2}}+y=0$
Sol :
y=5cosx-3sinx
Differentiating w.r.t x
$\frac{d y}{dx}=-5\sin x-3 \cos x$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-5 \cos x-3(-\sin x)$
$\frac{d^{2} y}{d x^{2}}=-[5\cos x-3 \sin x]$
$\frac{d^{2} y}{d x^{2}}=-y \Rightarrow \frac{d^{2} y}{d x^{2}}+y=0$
Question 9
A और B ज्ञात करे ताकि(Find A and B such that) y=Asin5x+Bcos5x समीकरण (satisfies the equation) $\frac{d^{2} y}{d x^{2}}+\frac{1}{5} \cdot \frac{d y}{d x}+15 y=101 \sin 5 x$ को संतुष्ट करता है ।
Sol :
y=Asin5x+Bcos5x
Differentiating w.r.t x
$\frac{d y}{d x}=5 A \cos x-5B \sin 5 x$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=-5 \times 5 A \sin 5 x-5 \times 5 B \cos 5 x$
=-25Asin5x-25Bcosx
∵$\frac{d^{2} y}{d x^{2}}+\frac{1}{5} \cdot \frac{d y}{d x}+15 y=101 \mathrm{sin} 5 x$
-25Asin5x-25Bcos5x+$\frac{1}{5}$(5Acos5x-5Bsin5x)+15(Asin5x+Bcos5x)=101sin5x
-25Asin5x-25Bcosx+Acos5x-Bsin5x+15Asin5x+15Bcos5x=101sin5x
-10Asin5x-Bsin5x-10Bcos5x+Acos5x=101sin5x
(-10A-B)sin5x+(-10B+A)cos5x=101sin5x+0.cos5x
By Equating
-10A-B=101..(i)×1
A-10B=0..(ii)×10
$\begin{aligned}-10A-B=101\\10A-100B=0 \\\hline -101B=101 \end{aligned}$
B=-1
समीकरण (ii) मे B का मान रखने पर ,
A-10(-1)=0
A=-10
Question 10
यदि(If) $y=3 e^{2 x}+2 e^{3 x}$ , सिद्द करे कि (prove that) $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$
Sol:
$y=3 e^{2 x}+2 e^{3 x}$
Differentiating w.r.t x
$\frac{d y}{d}=3 \cdot e^{2 x} \cdot 2+2 \cdot e^{3 x} \cdot 3$
$\frac{d y}{d x}=6 e^{2 x}+6 \cdot e^{3 x}$
Again ,Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=6 \cdot e^{2 x} \cdot 2+6 \cdot e^{3 x} \cdot 3$
$\frac{d^{2} y}{d x^{2}}=12 e^{2 x}+18 e^{3 x}$
$\frac{d^{2} y}{d x^{2}}=30 e^{2 x}+30 e^{3 x}-18 e^{2 x}-12 e^{3 x}$
$\frac{d^{2} y}{d x^{2}}=5\left(6 e^{2 x}+6 e^{3 x}\right)-6\left(3 e^{2 x}+2 e^{3 x}\right)$
$\frac{d^{2} y}{d x^{2}}=5 \frac{d y}{d x}-6 y$
$\frac{d^{2} y}{d x^{2}}-\frac{5 dy}{dx}+6 y=0$
Question 11
यदि(If) $y=\mathrm{Ae}^{mx}+\mathrm{Be}^{2x}$ , दिखाएँ कि (show that) $\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0$
Sol :
$y=\mathrm{Ae}^{mx}+\mathrm{Be}^{2x}$
Differentiating w.r.t x
$\frac{d y}{d x}=A e^{m x} \times m+B \cdot e^{n x} \cdot n$
$\frac{d y}{d x}=m A e^{m x}+n B e^{n x}$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=m A \cdot e^{m x} \cdot m+n B \cdot e^{n x} \cdot n$
$\frac{d^{2} y}{d x^{2}}=m^{2} A e^{m x}+n^{2} B e^{n x}$
$\frac{d^{2} y}{d x^{2}}=m^{2} A e^{m x}+m n B e^{n x}+m n A e^{m x}+n^2 Be^{nx}-mnAe^{mx}-mnBe^{nx}$
$\frac{d^{2} y}{d x^{2}}=m\left(m A e^{m x}+n B e^{n x}\right)+n\left(m A e^{m}+n B e^{nx}\right)-mn(Ae^{mx}+Be^{nx})$
$\frac{d^{2} y}{d x^{2}}=(m+n) \cdot\left(m A e^{m x}+n B e^{n x}\right)-m n\left(A e^{m x}+B e^{nx}\right)$
$\frac{d^{2} y}{d x^{2}}=(m+n) \frac{d y}{d x}-m n y$
$\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d}+m n y=0$
Question 12
यदि(If) $y=500 e^{7 x}+600 e^{-7 x}$ , दिखाएँ कि (show that ) $\frac{d^{2} y}{d x^{2}}=49 y$
Sol :
$y=500 e^{7 x}+600 e^{-7 x}$
Differentiating w.r.t x
$\frac{d y}{d x}=500 e^{7 x} \cdot 7+600 e^{-7 x} \cdot(-7)$
$\frac{d y}{d x}=7 \times 500 e^{7 x}-7 \times 600 e^{-7 x}$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=7 \times 500 e^{7 x} \times 7-7 \times 600 e^{-7x} \times(-7)$
$\frac{d^{2}y}{d^{2}x}=49\left[500 e^{7 x}+600 e^{-7x}\right]$
$\frac{d^{2} y}{d x^{2}}=49 y$
Question 13
यदि(If) $y=\sin ^{-1} x$ , दिखाएँ कि (show that) $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=0$
Sol :
$y=\sin ^{-1} x$
Differentiating w.r.t x
$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{d x^{2}}=\frac{0 \cdot \sqrt{1-x^{2}}-1 \cdot \frac{1}{2 \sqrt{1-x^{2}}} \times(-2x)}{(\sqrt{1-x^{2}})^{2}}$
$\frac{d^{2} y}{d x^{2}}=\frac{\frac{x}{\sqrt{1-x^{2}}}}{1-x^{2}}$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=\frac{x}{\sqrt{1-x^{2}}}$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x}=x \cdot \frac{d y}{d x}$
$\left(1-x^{2}\right) \frac{d^{2} y}{dx^{2}}-x \frac{d y}{d x}=0$
Question 14
यदि(If) $y=\left(\tan ^{-1} x\right)^{2}$ , दिखाएँ कि (show that) $\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}-2=0$
Sol :
$y=\left(\tan ^{-1} x\right)^{2}$
Differentiating w.r.t x
$y_{1}=2\left(\tan ^{-1} x\right) \times \frac{1}{1+x^{2}}$
$y_{1}=\frac{2 \tan ^{-1} x}{1+x^{2}}$
$2 \tan ^{-1} x=\left(1+x^{2}\right) y_{1}$
Again , Differentiating w.r.t x
$y_{2}=\frac{2 \times \frac{1}{1+x^{2}} \times\left(1-x^{2}\right)-2 \tan ^{-1} x \times 2 x}{\left(1+x^{2}\right)^{2}}$
$y_{2}=\frac{2-4 x \tan ^{-1} x}{\left(1+x^{2}\right)^{2}}$
$\left(1+x^{2}\right)^{2} \cdot y_{2}=2-2 x \cdot 2 \tan ^{-1} x$
$\left(x^{2}+1\right)^{2} y_{2}=2-2 x\left(1+x^{2}\right) y_{1}$
$\left(x^{2}+1\right)^{2} y_{2}+2 x\left(x^{2}+1\right) y_{1}-2=0$
Question 15
यदि (If) y=3cos(log x)+4sin(log x) दिखाएँ कि (show that) $x^{2} y_{2}+x y_{1}+y=0$
Sol :
y=3cos(log x)+4sin(log x)
Differentiating w.r.t x
$y_{1}=-3 \cdot \sin (\log x) \times \frac{1}{x}+4 \cos (\log x) \times \frac{1}{x}$
$y_{1}=\frac{1}{x}[-3 \sin (\log x)+4 \cos (\log x)]$
Again , Differentiating w.r.t x
$y_{2}=\frac{-1}{x^{2}}\left[-3 \sin\left(\log x\right)+4 \cos \left(\log x\right)\right]+\frac{1}{x}\left[\frac{-3cos(log x)}{x}+\frac{4\{-sin(\log x)\}}{x}\right]$
$y_{2}=-\frac{1}{x^{2}}\left[-3 \sin \left(\log {x}\right)+4 \cos (\log x)\right]+\frac{1}{x^{2}}\left[-3 \cos \left(\log x\right)-4sin(\log x)\right]$
$x^{2} y_{2}=-\left[-3 \sin \left(\log x\right)+4 \cos (\log x)\right]-[3cos(\log x)+4\sin (\log x)]$
$x^{2} y_{2}=-x y_{1}-y$
$x^{2} y_{2}+x y_{1}+y_{2}=0$
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
$\frac{d^{2} y}{d x^{2}}=\frac{-a \cdot e^{a \cos ^{-1} x}.a \cdot \frac{1}{\sqrt{1-x^{2}}}\times \sqrt{1-x^{2}}-\left.(-a e^{\left.a \cos ^{-1} x\right.}) \times \frac{1}{2 \sqrt{1-x^{2}}}\right.\times -2x}{(\sqrt{1-x^{2}})^{2}}$
$\frac{d^{2} y}{d x^{2}}=\frac{a^{2} e^{a \cos^{-1}}x-\frac{a x \cdot e^{a \cos^{-1}}x}{\sqrt{1-x^{2}}}}{1-x^{2}}$
$\left(1-x^{2}\right) \cdot \frac{d^{2} y}{d x^{2}}=a^{2} \cdot y+x \cdot \frac{d y}{d x}$
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-a^{2} y=0$
Question 19
यदि(If) $y=e^{\tan ^{-1} x}$ दिखाएँ कि(show that) $\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0$
Sol :
$y=e^{\tan ^{-1} x}$
Differentiating w.r.t x
$\frac{d y}{d x}=e^{\tan ^{-1} x}\times\frac{1}{1+x^{2}}$
$\frac{d y}{d x}=\frac{e^{\tan ^{-1} }x }{1+x^{2}}$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{dx^{2}}=\frac{e^{\tan ^{-1} x} \cdot \frac{x}{1+x^{2}} \times\left(1+x^{2}\right)-e^{\tan x} \cdot 2 x}{\left(1+x^{2}\right)^{2}}$
$\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}=-2 x e^{\tan ^{-1} x}+e^{\tan ^{-1} x}$
$\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}=-e^{\tan ^{-1} x}(2 x-1)$
$\left(1+x^{2}\right) \frac{d^{2} y}{dx^{2}}=-\frac{e^{\tan ^{-1} x}}{1+x^{2}}(2 x-1)$
$\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}=-(2 x-1) \frac{d y}{d x}$
$\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0$
Question 20
यदि(If) $y=\mathrm{A} e^{k t} \cos (p t+\alpha)$ , दिखाएँ कि(show that)
$\frac{d^{2} y}{d t^{2}}-2 k \frac{d y}{d t}+\left(p^{2}+k^{2}\right) y=0$
Sol :
$y=\mathrm{A} e^{k t} \cos (p t+\alpha)$
Differentiating w.r.t x
$\frac{d y}{d t}=A e^{k t} \times k \cdot \cos (p t+\alpha)+A e^{k t} \cdot\left[-\sin(p t+\alpha)\right] p$
$\frac{d y}{d t}=A k e^{k t} \cos (p t+\alpha)-A p e^{k t} \operatorname{sin}(p t+\alpha)$
Again , Differentiating w.r.t x
$\frac{d^{2} y}{d t^{2}}=A k e^{k t} \times k \cos (p t+\alpha)+A k e^{k t} \cdot\{-\sin (p t+x)\}p-Ape^{kt}\times k\sin(pt+\alpha)-Ape^{kt}.\cos(pt+\alpha).p$
$\frac{d^{2} y}{d t^{2}}=A k^{2} e^{k t} \cos (p t+\alpha)-A k p e^{k t} \sin (p t+\alpha)-A k p e^{k t} \sin (p t+\alpha)-A p^{2} e^{kt} \cos (p t+\alpha)$
$\frac{d^{2} y}{d t^{2}}=2 A k^{2} e^{k t} \cos (p t+t)-2 A k p e^{k t} \sin (p t+\alpha)-A k^{2} e^{k t} \cos (p t+\alpha)-A p^{2} e^{k t}\cos(pt+\alpha)$
$\frac{d^{2} y}{d t^{2}}=2 k\left[A k e^{k t} \cos(p t+\alpha)-A p e^{k t} \sin (p t+\alpha)\right]-\left(k^{2}+p^{2}\right) A e^{k+1} \cdot \cos(p+\alpha)$
$\frac{d^{2} y}{d t^{2}}=2 k \frac{d y}{d t}-\left(k^{2}+p^{2}\right) y$
$\frac{d^{2} y}{d t^{2}}-2 k \frac{d y}{d t}+\left(p^{2}+k^{2}\right) y=0$
Question 21
यदि(If) $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2 x$ , सिद्द करे कि(prove that) $\left(x^{2}-1\right) y_{2}+x y_{1}-m^{2} y=0$
Sol :
$y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2 x$
Differentiating w.r.t x
$\frac{1}{m} \cdot y^{\frac{1}{k}-1} \cdot \frac{d y}{dx}+\left(\frac{-1}{m}\right) \cdot y^{-\frac{1}{m}-1} \frac{dy}{d x}=2$
$\frac{1}{m} y^{-1}\left[y^{\frac{1}{m}}-y^{-\frac{1}{m}}\right] \frac{d y}{dx}=2$
$\frac{1}{m{y}}\left[y^{\frac{1}{m}}-y^{-\frac{1}{m}}\right] \frac{d y}{dx}=2$
$\left[y^{\frac{1}{m}}-y^{-\frac{1}{m}}\right] \frac{d y}{dx}=2 m y$
दोनो तरफ वर्ग करने पर ,
$\left[y^{\frac{1}{m}}-y^{-\frac{1}{m}}\right]^{2}\left(\frac{d y}{dx}\right)^{2}=4 m^{2} y^{2}$
$\left[y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2 \cdot y^{\frac{1}{m}} \cdot y^{-\frac{1}{m}}\right]\left(\frac{dy}{d x}\right)^{2}=4 m^{2} y^{2}$
$\left[y^{\frac{2}{m}}+y^{\frac{-2}{m}}-2\right]\left(\frac{dy}{dx}\right)^{2}=4 m^{2} y^{2}$..(i)
∵ $y^{\frac{1}{m}}+y^{-\frac{1}{m }}=2 x$
दोनो तरफ वर्ग करने पर ,
$y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2 \cdot y^{\frac{1}{m}} \cdot y^{-\frac{1}{m}}=4 x^{2}$
$y^{\frac{2}{m}}+y^{-\frac{2}{m}}=4 x^{2}-2$
∴$[4x^2-2-2]\left(\frac{dy}{dx}\right)^2=4m^2y^2$ समीकरण (i) से
$\left(x^{2}-1\right) \cdot\left(\frac{d y}{dx}\right)^{2}=m^{2} y^{2}$
Differentiating w.r.t x
$2 x \cdot\left(\frac{d y}{d x}\right)^{2}+\left(x^{2}-1\right) \cdot 2\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}=m^{2} \cdot 2 y \frac{d y}{dx}$
$2 \frac{d y}{d x}\left[x\left(\frac{d y}{d x}\right)+\left(x^{2}-1\right) \cdot \frac{d^{2} y}{dx^{2}}\right]=m^{2} \cdot 2 y \frac{d y}{dx}$
$\left(x^{2}-1\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{2}-m^{2} y=0$
$\left(x^{2}-1\right) \cdot y_{2}+x y_{1}-m^{2} y=0$