Question

Consider the differential equation: (y^2+2*x)dx+(2*x*y-1)dy = 0. (a) show that the equation is exact by evaluating (\partial m)/(\partial y) = (\partial n)/(\partial x)= (b) determine the general solution in implicit form: incorrect: your answer is incorrect. = c

Answer 1

The given ODE

$\underbrace{(y^2+2x)}_M\,\mathrm dx+\underbrace{(2xy-1)}_N\,\mathrm dy=0$

is exact if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. It so happens that we have $\frac{\partial M}{\partial y}=2y=\frac{\partial N}{\partial x}$, so it is indeed exact. For such an ODE, we're looking for a solution of the form $\Psi(x,y)=C$, for which the differential is

$\mathrm d\Psi=\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0$

so we have the following system of PDEs that allow us to solve for $\Psi$:

$\dfrac{\partial\Psi}{\partial x}=M=y^2+2x$

$\dfrac{\partial\Psi}{\partial y}=N=2xy-1$

In the first PDE, we can integrate both sides with respect to $x$ and recover $\Psi$:

$\displaystyle\int\frac{\partial\Psi}{\partial x}\,\mathrm dx=\int(y^2+2x)\,\mathrm dx$

$\implies\Psi(x,y)=xy^2+x^2+f(y)$

Then differentiating this with respect to $y$ returns $N$:

$\dfrac{\partial\Psi}{\partial y}=2xy+\dfrac{\mathrm df}{\mathrm dy}=2xy-1$

$\implies\dfrac{\mathrm df}{\mathrm dy}=-1$

$\implies f(y)=-y+C$

So the general solution to the ODE is

$\Psi(x,y)=xy^2+x^2-y+C=C$

or simply

$xy^2+x^2-y=C$