IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rahulraj1

e^2x-3ydx+e^2y-3xdy=0

raulrag009

here is my solution

$e^{2x-3y}dx = -e^{2y-3x}dy\\\\ \frac{dy}{dx}=-\frac{e^{2x-3y}}{e^{2y-3x}}\\\\ \frac{dy}{dx}=-e^{2x-3y-2y+3x}\\\\ \frac{dy}{dx}=-e^{5(x-y)}\\\\ Put\;x-y=t\\\\ 1-\frac{dy}{dx}=\frac{dt}{dx}\\\\ 1-\frac{dt}{dx}=-e^{5t}\\\\ \int{\frac{dt}{1+e^{5t}}}=\int{dx}\\\\ \int{\frac{e^{-5t}dt}{e^{-5t}+1}}=x+C\\\\ Put \;e^{-5t}+1=V\\\\ -5e^{-5t}dt=dv\\\\\\\\ -\frac{1}{5}\int{\frac{dv}{v}}=x+C\\\\ -\frac{1}{5}[\log{1+e^{-5(x-y)}}]=x+C$

forgive me for my calculation mistakes

sarvpriye

Solution is simple:

e^2x-3ydx + e^2y-3xdy =0

(e^2xdx)/(e^3y) = - (e^2ydy)/(e^3x)

e^5xdx= - e^5ydy

integrating both sides, we get

e^5x/5 + e^5y/5 = K

(k is any constant)

or simply

e^5x+ e^5y = C

(C=5K)

raulrag009

Damn , how could i miss such a simple one

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