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Varvara68 [4.7K]
2 years ago
8

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

. Integrate each term of this partial derivative with respect to x, letting h(y) be an
Mathematics
1 answer:
IRISSAK [1]2 years ago
5 0

(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0

Suppose the ODE has a solution of the form F(x,y)=C, with total differential

\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

This ODE is exact if the mixed partial derivatives are equal, i.e.

\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}

We have

\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)

\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)

so the ODE is indeed exact.

Integrating both sides of

\dfrac{\partial F}{\partial x}=(x+y)^2

with respect to x gives

F(x,y)=\dfrac{(x+y)^3}3+g(y)

Differentiating both sides with respect to y gives

\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2

\implies g(y)=-\dfrac{y^3}3-2y+C

\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C

so the general solution to the ODE is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C

Given that y(1)=1, we find

\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13

so that the solution to the IVP is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13

\implies\boxed{(x+y)^3-y^3-6y=1}

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