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Differential Equations question

2017 · Shift 2 · Q20
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  5. /2017 · Shift 2 · Q20

Differential Equations question

2017 · Shift 2 · Q20

JEE AdvancedMathematicsDifferential EquationsMCQ+3 / −1
If y = y(x) satisfies the differential equation 8x(9+x)dy=(4+9+x)−1{8\sqrt x \left( {\sqrt {9 + \sqrt x } } \right)dy = {{\left( {\sqrt {4 + \sqrt {9 + \sqrt x } } } \right)}^{ - 1}}}8x​(9+x​​)dy=(4+9+x​​​)−1 dx, x > 0 and y(0) =7\sqrt 77​, then y(256) =
  1. A
    16
  2. B
    3
  3. C
    9
  4. D
    80
View written solutionFree

Correct answer: B

  1. Interpret the differential equation

Given

8x 9+x dy=(4+9+x)−1dx,8\sqrt{x}\,\sqrt{9+\sqrt{x}}\,dy=\left(\sqrt{4+\sqrt{9+\sqrt{x}}}\right)^{-1}dx,8x​9+x​​dy=(4+9+x​​​)−1dx,

with x>0x>0x>0 and y(0)=7y(0)=\sqrt{7}y(0)=7​.

So,

dydx=18x9+x4+9+x.\frac{dy}{dx}=\frac{1}{8\sqrt{x}\sqrt{9+\sqrt{x}}\sqrt{4+\sqrt{9+\sqrt{x}}}}.dxdy​=8x​9+x​​4+9+x​​​1​.
  1. Recognize the chain-rule structure

Let

u=4+9+x.u=4+\sqrt{9+\sqrt{x}}.u=4+9+x​​.

Then

ddx(4+9+x)=124+9+x⋅ddx(9+x).\frac{d}{dx}\big(\sqrt{4+\sqrt{9+\sqrt{x}}}\big) =\frac{1}{2\sqrt{4+\sqrt{9+\sqrt{x}}}}\cdot \frac{d}{dx}\big(\sqrt{9+\sqrt{x}}\big).dxd​(4+9+x​​​)=24+9+x​​​1​⋅dxd​(9+x​​).

Now,

ddx(9+x)=129+x⋅ddx(x)=129+x⋅12x.\frac{d}{dx}\big(\sqrt{9+\sqrt{x}}\big) =\frac{1}{2\sqrt{9+\sqrt{x}}}\cdot \frac{d}{dx}(\sqrt{x}) =\frac{1}{2\sqrt{9+\sqrt{x}}}\cdot \frac{1}{2\sqrt{x}}.dxd​(9+x​​)=29+x​​1​⋅dxd​(x​)=29+x​​1​⋅2x​1​.

Hence,

ddx(4+9+x)=124+9+x⋅14x9+x=18x9+x4+9+x.\frac{d}{dx}\big(\sqrt{4+\sqrt{9+\sqrt{x}}}\big) =\frac{1}{2\sqrt{4+\sqrt{9+\sqrt{x}}}}\cdot \frac{1}{4\sqrt{x}\sqrt{9+\sqrt{x}}} =\frac{1}{8\sqrt{x}\sqrt{9+\sqrt{x}}\sqrt{4+\sqrt{9+\sqrt{x}}}}.dxd​(4+9+x​​​)=24+9+x​​​1​⋅4x​9+x​​1​=8x​9+x​​4+9+x​​​1​.

But this is exactly dydx\dfrac{dy}{dx}dxdy​. Therefore,

dydx=ddx(4+9+x).\frac{dy}{dx}=\frac{d}{dx}\big(\sqrt{4+\sqrt{9+\sqrt{x}}}\big).dxdy​=dxd​(4+9+x​​​).

So,

y=4+9+x+C.y=\sqrt{4+\sqrt{9+\sqrt{x}}}+C.y=4+9+x​​​+C.
  1. Use the initial condition

Given y(0)=7y(0)=\sqrt{7}y(0)=7​.

At x=0x=0x=0,

x=0,\sqrt{x}=0,x​=0,

so

y(0)=4+9+0+C=4+3+C=7+C.y(0)=\sqrt{4+\sqrt{9+0}}+C=\sqrt{4+3}+C=\sqrt{7}+C.y(0)=4+9+0​​+C=4+3​+C=7​+C.

Since y(0)=7y(0)=\sqrt{7}y(0)=7​,

7+C=7  ⟹  C=0.\sqrt{7}+C=\sqrt{7}\implies C=0.7​+C=7​⟹C=0.

Thus,

y=4+9+x.y=\sqrt{4+\sqrt{9+\sqrt{x}}}.y=4+9+x​​​.
  1. Find y(256)y(256)y(256)

Since

256=16,\sqrt{256}=16,256​=16,

we get

9+256=9+16=25=5.\sqrt{9+\sqrt{256}}=\sqrt{9+16}=\sqrt{25}=5.9+256​​=9+16​=25​=5.

Therefore,

y(256)=4+5=9=3.y(256)=\sqrt{4+5}=\sqrt{9}=3.y(256)=4+5​=9​=3.
  1. Check options
  • A: 161616 ❌
  • B: 333 ✅
  • C: 999 ❌
  • D: 808080 ❌

Hence, the correct option is B.

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