Question

Solve the given Differential Equation:
dy/dx +y cosx = (e^sinx)cosx

Answer 1

If you write it as dy/dx = cos x * y + e^sinx*cosx,
you'll see that it is a first order linear differential equation.

These are all of the form dy/dx=a(x)*y(x) + b(x).
The solution can be obtained from Leibniz' Formula:\[y(x)=e^{-\int\limits_{}^{}a(x)dx}\left( \int\limits_{}^{}e^{\int\limits_{}^{}a(x)dx}\cdot b(x)+C \right)\]I don't know if you have any experience with this formula, but it looks worse than it is ;)

Because a(x)=cos x, ∫a(x)dx is easily found. Give it a try, let me know how what happened!

Answer 2

Small typo: dy/dx = -cos x * y + e^sinx*cosx, so a(x) = -cosx.

Answer 3

I have got an I.F. as e^(integral of cos x)=e^sinx).
Now I put this value as y.(e^sin x) = integral of (( 2e^sinx).cosx)
Later when it came to int. part i did it "BY PARTS".. But the problem is that
reapperance of original int is found everytime after solving..How to cope this one Dude!!!

Answer 4

You don't need integration by parts, because everything comes out easier than it looks:

∫a(x)dx = ∫-cos(x)dx = -sinx.

Put these into Leibniz's Formula:\[y(x)=e^{\sin x}\left( \int\limits_{}^{}e^{-\sin x}\cdot e^{\sin x}\cos x dx + C \right)\]Now both e-powers in the middle cancel out, so you only have:\[y(x)=e^{\sin x}\left( \int\limits_{}^{}\cos x dx +C \right)=\]\[y(x)=e^{\sin x}(\sin x +C)\]

Answer 5

Thank you bro. I got it.

Answer 6

ZeHanz Can u give me your email-Id?

Answer 7

It is my OpenStudy name, at gmail dot com

Answer 8

Hmmm!!!

Answer 9

Dude I want a advice from you. I have my Mathematics Paper of Class XII(ISC BOARD) on 26th Feb. I wanna score Cent % in Maths. I have completed about 90% of my syllabus and having some problems in chapters like D.E.; Conics; Determinants.. Can you suggest me some tips 2 score cent!!

Answer 10

I'm not familiar with your syllabus. I think the only thing you can do is: practise more problems! Everytime you solve a problem, you'll get closer to your perfect score!