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Differential Equations question

2023 · 11 Apr · Shift 2 · Q34
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  5. /2023 · 11 Apr · Shift 2 · Q34

Differential Equations question

2023 · 11 Apr · Shift 2 · Q34

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dydx+5x(x5+1)y=(x5+1)2x7,x>0\frac{d y}{d x}+\frac{5}{x\left(x^{5}+1\right)} y=\frac{\left(x^{5}+1\right)^{2}}{x^{7}}, x \gt 0dxdy​+x(x5+1)5​y=x7(x5+1)2​,x>0. If y(1)=2y(1)=2y(1)=2, then y(2)y(2)y(2) is equal to :
  1. A
    693128\frac{693}{128}128693​
  2. B
    697128\frac{697}{128}128697​
  3. C
    637128\frac{637}{128}128637​
  4. D
    679128\frac{679}{128}128679​
View written solutionFree

Correct answer: A

  1. We have the linear differential equation
dydx+5x(x5+1)y=(x5+1)2x7,x>0\frac{dy}{dx}+\frac{5}{x(x^5+1)}y=\frac{(x^5+1)^2}{x^7}, \qquad x>0dxdy​+x(x5+1)5​y=x7(x5+1)2​,x>0

with initial condition

y(1)=2.y(1)=2.y(1)=2.

We solve it using the integrating factor method.

  1. Compare with the standard form
dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

where

P(x)=5x(x5+1),Q(x)=(x5+1)2x7.P(x)=\frac{5}{x(x^5+1)}, \qquad Q(x)=\frac{(x^5+1)^2}{x^7}.P(x)=x(x5+1)5​,Q(x)=x7(x5+1)2​.

The integrating factor is

I.F.=e∫P(x) dx=e∫5x(x5+1)dx.\text{I.F.}=e^{\int P(x)\,dx}=e^{\int \frac{5}{x(x^5+1)}dx}.I.F.=e∫P(x)dx=e∫x(x5+1)5​dx.

Now simplify the integrand:

5x(x5+1)=5x−5x4x5+1\frac{5}{x(x^5+1)}=\frac{5}{x}-\frac{5x^4}{x^5+1}x(x5+1)5​=x5​−x5+15x4​

since

5x−5x4x5+1=5(x5+1)−5x5x(x5+1)=5x(x5+1).\frac{5}{x}-\frac{5x^4}{x^5+1} =\frac{5(x^5+1)-5x^5}{x(x^5+1)} =\frac{5}{x(x^5+1)}.x5​−x5+15x4​=x(x5+1)5(x5+1)−5x5​=x(x5+1)5​.

Therefore,

∫5x(x5+1)dx=∫5xdx−∫5x4x5+1dx=5ln⁡x−ln⁡(x5+1).\int \frac{5}{x(x^5+1)}dx =\int \frac{5}{x}dx-\int \frac{5x^4}{x^5+1}dx =5\ln x-\ln(x^5+1).∫x(x5+1)5​dx=∫x5​dx−∫x5+15x4​dx=5lnx−ln(x5+1).

So,

I.F.=e5ln⁡x−ln⁡(x5+1)=x5x5+1.\text{I.F.}=e^{5\ln x-\ln(x^5+1)}=\frac{x^5}{x^5+1}.I.F.=e5lnx−ln(x5+1)=x5+1x5​.
  1. Multiply the differential equation by the integrating factor:
x5x5+1dydx+x5x5+1⋅5x(x5+1)y=x5x5+1⋅(x5+1)2x7.\frac{x^5}{x^5+1}\frac{dy}{dx}+\frac{x^5}{x^5+1}\cdot \frac{5}{x(x^5+1)}y =\frac{x^5}{x^5+1}\cdot \frac{(x^5+1)^2}{x^7}.x5+1x5​dxdy​+x5+1x5​⋅x(x5+1)5​y=x5+1x5​⋅x7(x5+1)2​.

The left side becomes

ddx(y⋅x5x5+1).\frac{d}{dx}\left(y\cdot \frac{x^5}{x^5+1}\right).dxd​(y⋅x5+1x5​).

The right side simplifies to

x5x5+1⋅(x5+1)2x7=x5+1x2=x3+1x2?\frac{x^5}{x^5+1}\cdot \frac{(x^5+1)^2}{x^7} =\frac{x^5+1}{x^2} =x^3+\frac{1}{x^2}?x5+1x5​⋅x7(x5+1)2​=x2x5+1​=x3+x21​?

Careful:

x5+1x2=x3+1x2\frac{x^5+1}{x^2}=x^3+\frac{1}{x^2}x2x5+1​=x3+x21​

is correct because

x5x2=x3,1x2=x−2.\frac{x^5}{x^2}=x^3, \qquad \frac{1}{x^2}=x^{-2}.x2x5​=x3,x21​=x−2.

Hence,

ddx(y⋅x5x5+1)=x3+1x2.\frac{d}{dx}\left(y\cdot \frac{x^5}{x^5+1}\right)=x^3+\frac{1}{x^2}.dxd​(y⋅x5+1x5​)=x3+x21​.
  1. Integrate both sides:
y⋅x5x5+1=∫(x3+1x2)dx=x44−1x+C.y\cdot \frac{x^5}{x^5+1}=\int \left(x^3+\frac{1}{x^2}\right)dx =\frac{x^4}{4}-\frac{1}{x}+C.y⋅x5+1x5​=∫(x3+x21​)dx=4x4​−x1​+C.

So,

y=x5+1x5(x44−1x+C).y=\frac{x^5+1}{x^5}\left(\frac{x^4}{4}-\frac{1}{x}+C\right).y=x5x5+1​(4x4​−x1​+C).
  1. Use the initial condition y(1)=2y(1)=2y(1)=2.

At x=1x=1x=1,

2=15+115(144−11+C)=2(14−1+C).2=\frac{1^5+1}{1^5}\left(\frac{1^4}{4}-\frac{1}{1}+C\right) =2\left(\frac14-1+C\right).2=1515+1​(414​−11​+C)=2(41​−1+C).

Thus,

1=14−1+C=−34+C1=\frac14-1+C=-\frac34+C1=41​−1+C=−43​+C

so

C=1+34=74.C=1+\frac34=\frac74.C=1+43​=47​.

Therefore,

y=x5+1x5(x44−1x+74).y=\frac{x^5+1}{x^5}\left(\frac{x^4}{4}-\frac{1}{x}+\frac74\right).y=x5x5+1​(4x4​−x1​+47​).
  1. Find y(2)y(2)y(2).

First compute the bracket:

244−12+74=164−12+74=4−12+74.\frac{2^4}{4}-\frac{1}{2}+\frac74 =\frac{16}{4}-\frac12+\frac74 =4-\frac12+\frac74.424​−21​+47​=416​−21​+47​=4−21​+47​.

Using denominator 444,

4=164,−12=−24,74=74.4=\frac{16}{4}, \qquad -\frac12=-\frac24, \qquad \frac74=\frac74.4=416​,−21​=−42​,47​=47​.

So,

4−12+74=16−2+74=214.4-\frac12+\frac74=\frac{16-2+7}{4}=\frac{21}{4}.4−21​+47​=416−2+7​=421​.

Next,

25+125=3332.\frac{2^5+1}{2^5}=\frac{33}{32}.2525+1​=3233​.

Hence,

y(2)=3332⋅214=693128.y(2)=\frac{33}{32}\cdot \frac{21}{4}=\frac{693}{128}.y(2)=3233​⋅421​=128693​.
  1. Compare with the options:
y(2)=693128\boxed{y(2)=\frac{693}{128}}y(2)=128693​​

So the correct option is A.

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