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  • Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e^x sin x.

15.    Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = e^x\sin x.

Answers (1)

best_answer

We first find the general solution of the given differential equation

Given,

y' = e^x\sin x

\\ \implies \int dy = \int e^x\sin xdx

\\ Let I = \int e^x\sin xdx \\ \implies I = \sin x.e^x - \int(\cos x. e^x)dx \\ \implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\ \implies 2I = e^x(\sin x - \cos x) \\ \implies I = \frac{1}{2}e^x(\sin x - \cos x)

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c

Now, Since the curve passes through (0,0)

y = 0 when x =0 

\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\ \implies c = \frac{1}{2}

Putting the value of c, we get:

\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\ \implies 2y -1 = e^x(\sin x - \cos x)

 

 

Posted by

HARSH KANKARIA

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