Question

solve the first order differential equation: y-x(dy/dx)=(dy/dx)y^2e^y

Answer 1

Rearrange the equation to put it into a standard form,\[y-(x^2+y^2e^y)\frac{dy}{dx}=0\]which is in the form, \[M(x,y)+N(x,y)\frac{dy}{dx}=0\]We check to see if it's exact. We need to ensure,\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} \rightarrow 1=1\]which is true. Now,

Answer 2

the potential function,\[\Psi(x,y)\]has that\[\frac{\partial \Psi}{\partial x}=M=y\]and\[\frac{\partial N}{\partial x}=N=-x^2-y^2e^y\]

Answer 3

Integrate the first partial:\[\Psi(x,y)=xy+c(y)\]and then take the partial derivative of this result with respect to y,\[\frac{\partial \Psi}{\partial y}=x+c'(y)\]so that we can identify it with \[\frac{\partial \Psi}{\partial y}=-x^2-y^2e^y\]

Answer 4

Above you were saying \[\frac{∂N}{∂x} = 1\] while indeed it is \[-2x\]

Answer 5

(note a mistake above dN/dx should be dPsi/dy)

Answer 6

I mean where you say "We check to see if it's exact."

Answer 7

Oh - thanks nowhereman - typo with x^2...it should just be x

Answer 8

right :-)

Answer 9

great - now the whole thing's polluted.

Answer 10

\[\frac{\partial \Psi}{\partial y}=-(x+ye^y)=x+c'(y)\]

Answer 11

so that\[c'(y)=-2x-ye^y \rightarrow c(y)=-2xy-e^y(y^2-2y+2)\]where the last integral is doable using a few integration by parts. Sub. this back in for our solution Psi to give,\[\Psi(x,y)=xy-2xy-e^y(y^2-2y+2)\]\[=-xy-e^y(y^2-2y+2)+c=K\]

Answer 12

where c is the constant of integration, and K is a constant due to the fact that this function Psi satisfies\[\frac{d \Psi(x,y)}{dx}=\frac{\partial \Psi}{\partial x}\frac{dx}{dx}+\frac{\partial \Psi}{\partial y}\frac{dy}{dx}=\Psi_{x}+\Psi_y \frac{dy}{dx}=0\]which is the form of the equation we had to solve in the beginning.

Answer 13

Are you seeing any mistakes, nowhereman?

Answer 14

So \[\Psi(x,y)=-xy-e^y(y^2-2y+2)=c\]for c, some constant.