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## User: Saurabh Singh

Reputation:
0
Status:
Trusted
Location:
cunning_chin
Last seen:
6 days, 8 hours ago
Joined:
2 months, 1 week ago
Email:
s*******@gmail.com

#### Posts by Saurabh Singh

<prev • 17 results • page 1 of 2 • next >
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...  $Put x^{\frac{1}{3} } = t \Longrightarrow x = t^3 \\$ $dx = 3t^2 dt \\$ $\int_0^{\infty} t^{\frac{3}{2}} e^{-t} . 3t^2 dt = 3 \int_0^{\infty}.e^{-t} . t^{\frac{7}{2}} dt \\$ $= 3 \Gamma{\frac{9}{2}} \\$ $= 3 * \frac{7}{2}* \frac{5}{2}* \frac{3}{2}* \frac{1}{2}* \Gamma{\frac{1} ... written 6 days ago by Saurabh Singh0 1 answer 55 views 1 answers ... $ 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 + ...
written 6 days ago by Saurabh Singh0
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...  $f(x,y) = \frac{y-x}{\sqrt{xy}}, x_0 =1, y_0 = 2 , h = 0.1 \\$ $At \thinspace x_1 = x_0 + h = 1 + 0.1 = 1.1 \\$ $At \thinspace y_1 = y_0 + hf(x_0,y_0) = 2 + (0.1)(0.7071) = 2.0707 \\$ $At \thinspace x_2 = x_1 + h = 1.1 + 0.1 = 1.1 \\$ $At \thinspace y_2 = y_1 + hf(x_1,y_1 ... written 6 days ago by Saurabh Singh0 1 answer 58 views 1 answers ... $ \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} \\ { ...
written 9 days ago by Saurabh Singh0
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...  $f(x,y) = \frac{y-x}{x}, x_0 =1, y_0 = 2 , h = 0.2 \\$ $At \thinspace x_1 = x_0 + h = 1 + 0.2 = 1.2 \\$ $At \thinspace y_1 = y_0 + hf(x_0,y_0) = 2 + 0.2(1) = 2.2 \\$ $At \thinspace x_2 = x_1 + h = 1.2 + 0.2 = 1.4 \\$ $At \thinspace y_2 = y_1 + hf(x_1,y_1) = 2.2 + 0.2(0.83 ... written 14 days ago by Saurabh Singh0 1 answer 59 views 1 answers ... $ \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 15 days ago by Saurabh Singh0
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...  $\frac{dy}{dx} = 3x + y^2, x_0 = 0, y_0 = 1 \\$ $\text { Taylor's Series is given by } \\$ $y = y_0 + (x-x_0)y_0'+ \frac{(x-x_0)^2} {2!}y_0'' \\$ $y' = 3x + y^2 \Longrightarrow = 3x_0 + y_0 ^2 = 3(0) + 1 = 1 \\$ $y'' = 3 + 2yy' \Longrightarrow y_0'' = 3 + 2y_0y_0' = 3 + 2(1)(1 ... written 15 days ago by Saurabh Singh0 1 answer 64 views 1 answers ...  ![][1]$ 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\theta}} * -2sin2\theta \\ $... written 19 days ago by Saurabh Singh0 1 answer 70 views 1 answers ...  ![][1]$ \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 \\  = 2a\int_0^{\pi} \sqrt ...
written 20 days ago by Saurabh Singh0
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...  $x = a(\theta - cos\theta) \Longrightarrow \frac{dx}{d\theta} = a(1-cos\theta) \\ y = a(1-cos\theta ) \Longrightarrow \frac{dy}{d\theta} = asin\theta \\$ $S = \int_0^{2\pi} \sqrt{ a^2(1-cos\theta)^2 + a^2 sin^2\theta} d\theta \\$ \$ S = a\int_0^{2\pi} \sqrt{ 1 + cos^2\theta - 2 cos\ ...
written 20 days ago by Saurabh Singh0

#### Latest awards to Saurabh Singh

Centurion 5 weeks ago, created 100 posts.
Rising Star 6 weeks ago, created 50 posts within first three months of joining.

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