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weeeeeb [17]
3 years ago
8

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
Mathematics
1 answer:
ruslelena [56]3 years ago
4 0

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

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