Find the solution of D.E Xe(x(2)+y) dx= y dy
you haven't written this properly...post again more carefuly :D
Yeah, I can't read this.
X Exp(x2+y) = y dy
\[xe^{x^2+y} dx=ydy?\]
sorry i'm new >> yeah that is right please help me
That's okay, you will learn. Back to the DE in our hands. This DE is separable since we can write it as \(x(e^{x^2}e^y)dx=y dy.\). Divide both sides by \(e^y\) and then integrate both sides.
Dividing both sides by \(e^y\) gives \[xe^{x^2}dx=\frac{y}{e^y}dy \implies \int xe^{x^2}dx=\int ye^{-y}dy.\]
Do you know how to evaluate the integrals?
what (frac)
What do you mean?
Hmm looks like your browser has some problems with showing latex. Anyways frac{x}{y} means x over y.
what about Homogeneous ?
What about it?
can solve with it ? is that Homogeneous Eq.?
It doesn't look homogeneous to me.
if it ? how i solve in this way? and can you help me to get me book with maths Eq.
Are you asking about how to solve Homogeneous DE's?
These are good notes on HDE http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx
thanks a lot ..... <3
Glad to help.
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