If y-yx satisfies the differential equation8 -x-9-xdy-4-9-x -1dx- x-0and y0-7- then y256-

Name: JEE- Grade 12- Mathematics- Differential Equation- Session 05- D10
Uploaded: 2022-07-04T10:36:42+05:30
Description: Dy=[ 18 √(x)(√(9+√(x))) (√(4+√(9+√(x))))]dx Let 4+√(9+√(x))=t ⇒ nbsp;12 (√(9+√(x))). 12 √(x)dx=dt ∴ dy=12 √(t)dt ⇒ y=√(t) +C [ after integrating both si ...

Dy=[ 18 √(x)(√(9+√(x))) (√(4+√(9+√(x))))]dx Let 4+√(9+√(x))=t ⇒ nbsp;12 (√(9+√(x))). 12 √(x)dx=dt ∴ dy=12 √(t)dt ⇒ y=√(t) +C [ after integrating both si ...

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Standard XII

Mathematics

Geometrical Applications of Differential Equations

If y=yx satis...

Question

If $y = y (x)$ satisfies the differential equation
$8 \sqrt{x} (\sqrt{9 + \sqrt{x}}) d y = {(\sqrt{4 + \sqrt{9 + \sqrt{x}}})}^{- 1} d x, x > 0$
and $y (0) = \sqrt{7},$ then $y (256) =$

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Solution

$d y = ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{1}{8 \sqrt{x} (\sqrt{9 + \sqrt{x}}) (\sqrt{4 + \sqrt{9 + \sqrt{x}}})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ d x Let 4 + \sqrt{9 + \sqrt{x}} = t \Rightarrow \frac{1}{2 (\sqrt{9 + \sqrt{x}})} . \frac{1}{2 \sqrt{x}} d x = d t ∴ d y = \frac{1}{2 \sqrt{t}} d t \Rightarrow y = \sqrt{t} + C [after integrating both sides] \Rightarrow y = \sqrt{4 + \sqrt{9 + \sqrt{x}}} + C y (0) = \sqrt{7} \Rightarrow C = 0 \Rightarrow y = \sqrt{4 + \sqrt{9 + \sqrt{x}}} \Rightarrow y (256) = 3$

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Geometrical Applications of Differential Equations

Standard XII Mathematics

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If y=y(x) satisfies the differential equation 8√x(√9+√x)dy=(√4+√9+√x)−1dx, x>0 and y(0)=√7, then y(256)=

dy=⎡⎢ ⎢ ⎢ ⎢⎣18√x(√9+√x)(√4+√9+√x)⎤⎥ ⎥ ⎥ ⎥⎦dxLet 4+√9+√x=t⇒12(√9+√x).12√xdx=dt∴dy=12√tdt⇒y=√t+C [ after integrating both sides]⇒y=√4+√9+√x+Cy(0)=√7⇒C=0⇒y=√4+√9+√x⇒y(256)=3

If $y = y (x)$ satisfies the differential equation
$8 \sqrt{x} (\sqrt{9 + \sqrt{x}}) d y = {(\sqrt{4 + \sqrt{9 + \sqrt{x}}})}^{- 1} d x, x > 0$
and $y (0) = \sqrt{7},$ then $y (256) =$