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Solution to the Exact Differential Equation 📂Odinary Differential Equations

Solution to the Exact Differential Equation

Solution

The solution to the given exact differential equation M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y)\dfrac{dy}{dx}=0 is as follows.

  • Step 0.

    Since the differential equation given is exact, an ψ\psi exists such that ψx=M,  ψy=N,  ψ=c\psi_{x}=M,\ \ \psi_{y}=N, \ \ \psi=c.

  • Step 1.

    Integrate ψx\psi_{x}. Then, differentiate the obtained ψ\psi with respect to yy to find h′(y)h^\prime(y). The entire process including Step 1 can be done the opposite way with respect to xx and yy as well.

    ψ=∫ψx=∫M(x,y)dx+h(y)  ⟹  ψy=∂∂y(∫M(x,y)dx)+h′(y)=N(x,y)  ⟹  h′(y)=N(x,y)−∂∂y(∫M(x,y)dx) \begin{align} && \psi &= \int \psi_{x} = \int M(x,y)dx + h(y) \label{eq1} \\ \implies && \psi_{y}&=\dfrac{\partial}{\partial y} \left( \int M(x,y)dx \right) +h^\prime(y)=N(x,y) \nonumber \\ \implies && h^\prime(y)&=N(x,y) - \dfrac{\partial}{\partial y} \left( \int M(x,y)dx \right) \nonumber \end{align}

  • Step 2.

    Integrate the found h′(y)h^\prime(y) to obtain h(y)h(y) and then substitute it into (eq1)\eqref{eq1}.

    h(y)=∫h′(y)dy=∫N(x,y)dy−∫[∂∂y(∫M(x,y)dx)]dy  ⟹  ψ(x,y)=∫M(x,y)dx+∫N(x,y)dy−∫[∂∂y(∫M(x,y)dx)]dy \begin{align*} && h(y)&=\int h^\prime(y) dy =\int N(x,y) dy - \int \left[ \dfrac{\partial}{\partial y} \left( \int M(x,y)dx \right) \right] dy \\ \implies && \psi (x,y)&= \int M(x,y)dx + \int N(x,y) dy - \int \left[ \dfrac{\partial}{\partial y} \left( \int M(x,y)dx \right) \right] dy \end{align*}

  • Step 3.

  • Finally, representing ψ\psi in an implicit form will be the solution to the differential equation.

    ∫M(x,y)dx+∫N(x,y)dy−∫[∂∂y(∫M(x,y)dx)]dy=c \int M(x,y)dx + \int N(x,y) dy - \int \left[ \dfrac{\partial}{\partial y} \left( \int M(x,y)dx \right) \right] dy=c

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Examples

1

Solve the differential equation (2x3y+4xy)+(12x4+2x2+3y2)y′=0(2x^3y+4xy)+(\frac{1}{2}x^4+2x^2+3y^2)y^\prime=0.

M(x,y)=2x3y+4xyN(x,y)=12x4+2x2+3y2 M(x,y)=2x^3y+4xy\quad N(x,y)=\frac{1}{2}x^4+2x^2+3y^2

Then,

My=2x3+4x=Nx M_{y}=2x^3+4x=N_{x}

which means the given differential equation is exact. Therefore,

ψx=M,ψy=N \psi_{x}=M,\quad \psi_{y}=N

an ψ=c\psi=c that satisfies it exists.

ψ=∫Mdx=∫(2x3y+4xy)dx=12x4y+2x2y+h(y)  ⟹  ψy=12x4+2x2+h′(y)=N(x,y)=12x4+2x2+3y2  ⟹  h′(y)=3y2  ⟹  h(y)=∫3y2dy=y3  ⟹  ψ=12x4y+2x2y+h(y)=12x4y+2x2y+y3 \begin{align*} && \psi&=\int M dx = \int (2x^3y+4xy) dx = \frac{1}{2}x^4y+2x^2y+h(y) \\ \implies && \psi_{y}&=\frac{1}{2}x^4+2x^2+h^\prime (y)=N(x,y)=\frac{1}{2}x^4+2x^2+3y^2 \\ \implies && h^\prime (y)&=3y^2 \\ \implies && h(y)&=\int 3y^2 dy = y^3 \\ \implies && \psi& =\frac{1}{2}x^4y+2x^2y+h(y) = \frac{1}{2}x^4y+2x^2y+y^3 \end{align*}

Hence, the general solution is the implicit form of 12x4y+2x2y+y3=c\frac{1}{2}x^4y+2x^2y+y^3=c.

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2

Solve the differential equation (3xy+y2)+(x2+xy)y′=0(3xy+y^2) + (x^2+xy)y^\prime =0.

M(x,y)=3xy+y2,  N(x,y)=x2+xy M(x,y)=3xy+y^2,\ \ N(x,y)=x^2+xy

Then,

My=3x+2y≠2x+y=Nx M_{y}=3x+2y \ne 2x+y=N_{x}

which means the given differential equation is not exact. Therefore, the method of solving exact differential equations cannot be used. Let’s check if it really doesn’t work. Though it’s not an exact differential equation,

ψx=M(x,y)=3xy+y2,ψy=N(x,y)=x2+xy \psi_{x}=M(x,y)=3xy+y^2,\quad \psi_{y}=N(x,y)=x^2+xy

Let’s assume that an ψ\psi that satisfies it exists. Following the method of solving exact differential equations,

ψ=∫ψxdx=∫3xy+y2dx=32x2y+xy2+h(y)ψy=32x2+2xy+h′(y)=N(x,y)=x2+xy  ⟹  h′(y)=−12x2−xy \begin{align*} && \psi& =\int \psi_{x} dx = \int 3xy+y^2 dx = \frac{3}{2}x^2y+xy^2+h(y) \\ && \psi_{y}&=\frac{3}{2}x^2 + 2xy+h^\prime (y)=N(x,y)=x^2+xy \\ \implies && h^\prime (y) &= -\frac{1}{2}x^2-xy \end{align*}

This result contradicts the fact that h′(y)h^\prime(y) is a function that depends only on yy. Therefore, an ψ(x,y)\psi (x,y) that satisfies the definition of an exact differential equation does not exist.

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