Exercise 12.1
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
$2 x \cdot\left(\frac{d y}{d x}\right)^{2}+\left(x^{2}-1\right) \cdot 2\left(\frac{d y}{d x}\right) \frac{d^{2} y}{d x^{2}}=m^{2} \cdot 2 y \frac{d y}{dx}$
$2 \frac{d y}{d x}\left[x\left(\frac{d y}{d x}\right)+\left(x^{2}-1\right) \cdot \frac{d^{2} y}{dx^{2}}\right]=m^{2} \cdot 2 y \frac{d y}{dx}$
$\left(x^{2}-1\right) \cdot y_{2}+x y_{1}-m^{2} y=0$
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