JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 5)
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 :
$${{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos x$$ such that y(0) = 7. Then y($$\pi$$) is equal to :
$$2{e^{{\pi ^2}}} + 5$$
$${e^{{\pi ^2}}} + 5$$
$$3{e^{{\pi ^2}}} + 5$$
$$7{e^{{\pi ^2}}} + 5$$
Explanation
$${{dy} \over {dx}} - 2xy = 2(2\sin x - 5)x - 2\cos x$$
IF = $${e^{ - {x^2}}}$$
So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx} $$
$$ \Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$
$$ \Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$
Given at x = 0, y = 7
$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2
So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$
Now, at x = $$\pi$$,
y = 5 + 2$${e^{{x^2}}}$$
IF = $${e^{ - {x^2}}}$$
So, $$y.{e^{ - {x^2}}} = \int {{e^{ - {x^2}}}(2x(2\sin x - 5) - 2\cos x)dx} $$
$$ \Rightarrow y.{e^{ - {x^2}}} = {e^{ - {x^2}}}(5 - 2\sin x) + c$$
$$ \Rightarrow y = 5 - 2\sin x + c.{e^{{x^2}}}$$
Given at x = 0, y = 7
$$\Rightarrow$$ 7 = 5 + c $$\Rightarrow$$ c = 2
So, $$y = 5 - 2\sin x + 2{e^{{x^2}}}$$
Now, at x = $$\pi$$,
y = 5 + 2$${e^{{x^2}}}$$
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