Medium MCQ +4 / -1 PYQ · JEE Mains 2021

Let y = y(x) be the solution of the differential equation

${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$ such that y(0) = 7. Then y($\pi$) is equal to :

  1. A $2{e^{{\pi ^2}}} + 5$ Correct answer
  2. B ${e^{{\pi ^2}}} + 5$
  3. C $3{e^{{\pi ^2}}} + 5$
  4. D $7{e^{{\pi ^2}}} + 5$

Solution

${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$<br><br>IF = ${e^{ - {x^2}}}$<br><br>So, $y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx}$<br><br>$\Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$<br><br>$\Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$<br><br>Given at x = 0, y = 7<br><br>$\Rightarrow$ 7 = 5 + c $\Rightarrow$ c = 2<br><br>So, $y = 5 - 2\sin x + 2{e^{{x^2}}}$<br><br>Now, at x = $\pi$,<br><br>y = 5 + 2${e^{{x^2}}}$

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Linear Differential Equations

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