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  • Let be the solution of the differential equation <img alt="\frac{\mathrm{d} y}{\mathrm{~d} x}=2(y+2 \sin x-5) x-2 \cos x" src="https://entrancecorner.codecogs.com/gif.la

Let y=y(x) be the solution of the differential equation \frac{\mathrm{d} y}{\mathrm{~d} x}=2(y+2 \sin x-5) x-2 \cos x such that y(0)=7. Then y(\pi) is equal to:
Option: 1 7 e^{\pi^{2}}+5
Option: 2 7 e^{\pi^{2}}+5
Option: 3 7 e^{\pi^{2}}+5
Option: 4 7 e^{\pi^{2}}+5
Option: 5 e^{\pi^{2}}+5
Option: 6 e^{\pi^{2}}+5
Option: 7 e^{\pi^{2}}+5
Option: 8 e^{\pi^{2}}+5
Option: 9 2 e^{\pi^{2}}+5
Option: 10 2 e^{\pi^{2}}+5
Option: 11 2 e^{\pi^{2}}+5
Option: 12 2 e^{\pi^{2}}+5
Option: 13 3 e^{\pi^{2}}+5
Option: 14 3 e^{\pi^{2}}+5
Option: 15 3 e^{\pi^{2}}+5
Option: 16 3 e^{\pi^{2}}+5

Answers (1)

best_answer

\frac{dy}{dx}-2xy= -10x+4x\sin x-2\cos x
I\; \; F= e^{\int -2xdx}= e^{-x^{2}}
y\cdot e^{-x^{2}}= -\int 10x\cdot e^{-x^{2}}+\int 4xe^{-x^{2}}\sin xdx-\int 2\cos x\cdot e^{-x^{2}}dx
-x^{2}= t\; and using by parts in second integral by taking 2x\cdot e^{-x^{2}} as second function.
y\cdot e^{-x^{2}}= 5e^{-x^{2}}-2\sin x\cdot e^{-x^{2}}+\int 2\cos xe^{-x^{2}}dx-\int 2\cos xe^{-x^{2}}dx+C
\Rightarrow y\cdot e^{-x^{2}}= 5e^{-x^{2}}-2\sin x\cdot e^{-x^{2}}+C
at x = 0, y = 7
\Rightarrow 7= 5+C\Rightarrow C= 2
y= 5-2\sin x+2 e^{x^{2}}
y\left ( \pi \right )= 5+2e^{\pi ^{2}}
Option(3)

Posted by

Kuldeep Maurya

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