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User: Saurabh Singh

Reputation:
0
Status:
Trusted
Location:
Vadodara
Twitter:
cunning_chin
Last seen:
2 months, 4 weeks ago
Joined:
5 months, 1 week ago
Email:
s*******@gmail.com

Posts by Saurabh Singh

<prev • 18 results • page 1 of 2 • next >
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Answer: A: Prove that $\beta(m,m)=2^{1-2m}\,\beta\left(m,\frac{1}{2}\right)(L)$
... $ \beta{(m,m)} = \frac{\Gamma{m}\Gamma{m}}{\Gamma{2m}} \\ $ $ \text{ By Duplication Formula,} \\ $ $ \Gamma{m} \Gamma{(m+ \frac{1}{2})} = \frac{\sqrt \pi}{2^{2m-1}} \sqrt{2m} \\ $ $ \frac{\Gamma{m}}{\Gamma{2m}} = \frac{\sqrt \pi}{2^{2m-1}} . \frac{1}{\Gamma{(m+ \frac{1}{2}})} \\ $ $ \beta{(m,m ...
written 12 weeks ago by Saurabh Singh0 • updated 10 weeks ago by awari.swati831160
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Answer: A: Show that $\int_{0}^{\infty}\,\sqrt{x}\,e^{-\sqrt[3]{x}}\,dx=\frac{315}{16}\sqrt
... $$ $$ $ Put \thinspace x^{ \frac{1}{3} } = t \Longrightarrow x = t^3 \\ $ $ dx = 3t^2 dt \\ $ $ \int_0^{\infty} t^{\frac{3}{2}} e^{-t} . 3t^2 dt = 3 \int_0^{\infty}.e^{-t} . t^{\frac{7}{2}} dt \\ $ $ = 3 \Gamma{\frac{9}{2}} \\ $ $ = 3 * \frac{7}{2}* \frac{5}{2}* \frac{3}{2}* \frac{1}{2}* \Ga ...
written 3 months ago by Saurabh Singh0
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Answer: A: Compute $y$ at $x=0.25$ by modified Euler's method , given $y'=2xy , y(0)=1$ .'
... $ f(x,y) = 2xy, \thinspace x_0 =1, y_0 = 1 , h = 0.25 \\ $ $ At \thinspace x_1 = x_0 + h = 0 + 0.25 = 0.25 \\ $ $ At \thinspace y_1 = y_0 + hf(x_0,y_0) = 1 + (0.25) . 2(0) = 1 \\ $ $ y_1 ^{(1)} = y_0 + \frac{h}{2} [ f(x_0,y_0) + f(x_1,y_1) ] \\ $ $ = y_0 + \frac{h}{2} [ 2 x_0 y_0 + 2x_1 ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Using Euler's method find the approximate value of y where when x=1.5 in five st
... $ f(x,y) = \frac{y-x}{\sqrt{xy}}, x_0 =1, y_0 = 2 , h = 0.1 \\ $ $ At \thinspace x_1 = x_0 + h = 1 + 0.1 = 1.1 \\ $ $ At \thinspace y_1 = y_0 + hf(x_0,y_0) = 2 + (0.1)(0.7071) = 2.0707 \\ $ $ At \thinspace x_2 = x_1 + h = 1.1 + 0.1 = 1.1 \\ $ $ At \thinspace y_2 = y_1 + hf(x_1,y_1) = 2. ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Solve $(D^3+D)y=cosec(x)$ by the method of variation of parameters.
... $ \text{ The Auxillary equation is } D^3 + D = 0 \Longrightarrow D = 0, \pm i \\ $ $ \text{ C.F. is } y_c = c_1 + c_2cosx + c_3sinx \\ $ $ \text {Let }y_p = u + vcosx + wsinx = uy_1 + vy_2 + wy_3 \\ $ $ W = \begin{vmatrix} y_1 & y_2 & y_3 \\ {y'_1} & {y'_2} & {y'_3} \\ {y''_1 ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Using Euler's method find the approximate value of y where $\frac{dy}{dx}=\frac
... $ f(x,y) = \frac{y-x}{x}, x_0 =1, y_0 = 2 , h = 0.2 \\ $ $ At \thinspace x_1 = x_0 + h = 1 + 0.2 = 1.2 \\ $ $ At \thinspace y_1 = y_0 + hf(x_0,y_0) = 2 + 0.2(1) = 2.2 \\ $ $ At \thinspace x_2 = x_1 + h = 1.2 + 0.2 = 1.4 \\ $ $ At \thinspace y_2 = y_1 + hf(x_1,y_1) = 2.2 + 0.2(0.8333) = ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Using Taylor's series expansion solve $\frac{dy}{dx}=-xy^2$ with x=0,y=2 and com
... $ \frac{dy}{dx} = -xy^2, x_0 = 0, y_0 = 2 \\ $ $ \text { Taylor's Series is given by } \\ $ $ y = y_0 + (x-x_0)y_0'+ \frac{(x-x_0)^2} {2!}y_0'' \\ $ $ y' = -xy^2 \Longrightarrow y_0' = -x_0y_0 ^2 = 0 \\ $ $ y'' = -x2yy' + y^2 \Longrightarrow y_0''' = -2x_0y_0y_0' - y_0^2 = -4 \\ $ $ y''' = ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Using Taylor's series method, obtain the solution of the differential equation
... $ \frac{dy}{dx} = 3x + y^2, x_0 = 0, y_0 = 1 $ $ \text { Taylor's Series is given by } $ $ y = y_0 + (x-x_0)y_0'+ \frac{(x-x_0)^2} {2!}y_0'' $ $ y' = 3x + y^2 \Longrightarrow = 3x_0 + y_0 ^2 = 3(0) + 1 = 1 $ $ y'' = 3 + 2yy' \Longrightarrow y_0'' = 3 + 2y_0y_0' = 3 + 2(1)(1) = 5 $ $ y''' ...
written 3 months ago by Saurabh Singh0 • updated 11 weeks ago by awari.swati831160
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Answer: A: Show that the perimeter of the lemiscate $r^2=a^2\, cos2\theta$ is $\frac{a}{\sq
... $$ $$ ![enter image description here][1] $ S = 4 \int_0^{\frac{\pi}{4} } \sqrt { r^2 + (\frac{dr}{d\theta})^2 } d\theta \\ $ $ = 4 \int_0^{\frac{\pi}{4} } \sqrt { a^2cos2\theta + (\frac{dr}{d\theta})^2 } d\theta \\ $ $ r = a\sqrt{cos2\theta } \\ $ $ \frac{dr}{d\theta} = a \frac{1}{2\sqrt{cos2\ ...
written 3 months ago by Saurabh Singh0 • updated 10 weeks ago by Sanket Shingote ♦♦ 220
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Answer: A: Show that the whole perimeter of the cardioide $r=a(1+cos\theta)$ is $89$. Also
... $$ $$ ![enter image description here][1] $ \text {Length of Cardiode =} 2 \int_0^{\pi} \sqrt {r^2 +(\frac{dr}{d\theta})^2 } d\theta \\ $ $ = 2\int_0^{\pi} \sqrt{ a^2(1+cos\theta)^2 + a^2 sin^2\theta} d\theta \\ $ $ = 2a\int_0^{\pi} \sqrt{ 1 + cos^2\theta +2 cos\theta + sin^2\theta} d\theta \\ ...
written 3 months ago by Saurabh Singh0 • updated 10 weeks ago by Sanket Shingote ♦♦ 220

Latest awards to Saurabh Singh

Centurion 4 months ago, created 100 posts.
Rising Star 4 months ago, created 50 posts within first three months of joining.


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