Question

(x^2y+e^x)dx-x^2dy=0

Answer 1

It looks like the differential equation is

`\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0`

Check for exactness:

`(\partial\left(x^2y+e^x\right))/(\partial y) = x^2 \\\\ (\partial\left(-x^2\right))/(\partial x) = -2x`

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

`\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0`

*is* exact. If this modified DE is exact, then

`(\partial\left(\mu\left(x^2y+e^x\right)\right))/(\partial y) = (\partial\left(-\mu x^2\right))/(\partial x)`

We have

`(\partial\left(\mu\left(x^2y+e^x\right)\right))/(\partial y) = \left(x^2y+e^x\right)(\partial\mu)/(\partial y) + x^2\mu \\\\ (\partial\left(-\mu x^2\right))/(\partial x) = -x^2(\partial\mu)/(\partial x) - 2x\mu \\\\ \implies \left(x^2y+e^x\right)(\partial\mu)/(\partial y) + x^2\mu = -x^2(\partial\mu)/(\partial x) - 2x\mu`

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

`x^2\mu = -x^2(\mathrm d\mu)/(\mathrm dx) - 2x\mu \\\\ (x^2+2x)\mu = -x^2(\mathrm d\mu)/(\mathrm dx) \\\\ (\mathrm d\mu)/(\mu) = -(x^2+2x)/(x^2)\,\mathrm dx \\\\ (\mathrm d\mu)/(\mu) = \left(-1-\frac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^(-x-2\ln|x|) = (e^(-x))/(x^2)`

The modified DE,

`\left(e^(-x)y + \frac1{x^2}\right) \,\mathrm dx - e^(-x)\,\mathrm dy = 0`

is now exact:

`\frac{\partial\left(e^(-x)y+\frac1{x^2}\right)}{\partial y} = e^(-x) \\\\ (\partial\left(-e^(-x)\right))/(\partial x) = e^(-x)`

So we look for a solution of the form F(x, y) = C. This solution is such that

`(\partial F)/(\partial x) = e^(-x)y + \frac1{x^2} \\\\ (\partial F)/(\partial y) = e^(-x)`

Integrate both sides of the first condition with respect to x :

`F(x,y) = -e^(-x)y - \frac1x + g(y)`

Differentiate both sides of this with respect to y :

`(\partial F)/(\partial y) = -e^(-x)+(\mathrm dg)/(\mathrm dy) = e^(-x) \\\\ \implies (\mathrm dg)/(\mathrm dy) = 0 \implies g(y) = C`

Then the general solution to the DE is

`F(x,y) = \boxed{-e^(-x)y-\frac1x = C}`