Question

EXACT DIFFERENTIAL EQUATIONS

Test for exactness and find the general solution:

1. (2x + 4y − 5)dx + (6y + 4x − 1)dy = 0

2. (y + 2xy³)dx + (1+ 3x²y² + x)dy = 0

3. 3x²ydx + (x³ + 2y⁴)dy = 0

4. (1/y)dx - (x/y²⁾dy = 0

5. (4x³ − y³)dx − 3xy²dy = 0

Answer 1

1.

$$(2x + 4y − 5)dx + (6y + 4x − 1)dy = 0$$

$$\dfrac{\partial Q}{\partial x}=4, \dfrac{\partial P}{\partial y}=4$$

The given equation is exact because the partial derivatives are the same:

$$\dfrac{\partial Q}{\partial x}=4=\dfrac{\partial P}{\partial y}$$

We have the following system of differential equations to find the function $u(x, y)$

$$\begin{cases} \dfrac{\partial u}{\partial x}=2x + 4y − 5 \\ \\ \dfrac{\partial u}{\partial y}=6y + 4x − 1 \end{cases}$$

$$u(x, y)=\int (2x + 4y − 5)dx=x^2+4xy-5x+\varphi (y)$$

$$\dfrac{\partial u}{\partial y}=4x+\varphi' (y)=6y + 4x − 1$$

$$\varphi' (y)=6y − 1$$

The general solution of the exact differential equation is given by

$$x^2+4xy-5x+3y^2-y=C$$

where $C$ is an arbitrary constant.

2.

$$(y + 2xy^3)dx + (1+ 3x^2y^2 + x)dy = 0$$

$$\dfrac{\partial Q}{\partial x}=6xy^2+1, \dfrac{\partial P}{\partial y}=1+6xy^2$$

The given equation is exact because the partial derivatives are the same:

$$\dfrac{\partial Q}{\partial x}=6xy^2+1=\dfrac{\partial P}{\partial y}$$

We have the following system of differential equations to find the function $u(x, y)$

$$\begin{cases} \dfrac{\partial u}{\partial x}=y + 2xy^3 \\ \\ \dfrac{\partial u}{\partial y}=1+ 3x^2y^2 + x \end{cases}$$

$$u(x, y)=\int (y + 2xy^3)dx=xy+x^2y^3+\varphi (y)$$

$$\dfrac{\partial u}{\partial y}=x+3x^2y^2+\varphi' (y)=1+ 3x^2y^2 + x$$

$$\varphi' (y)= 1$$

The general solution of the exact differential equation is given by

$$xy+x^2y^3+y=C$$

where $C$ is an arbitrary constant.

3.

$$3x^2ydx + (x^3+2y^4)dy = 0$$

$$\dfrac{\partial Q}{\partial x}=3x^2, \dfrac{\partial P}{\partial y}=3x^2$$

The given equation is exact because the partial derivatives are the same:

$$\dfrac{\partial Q}{\partial x}=3x^2=\dfrac{\partial P}{\partial y}$$

We have the following system of differential equations to find the function $u(x, y)$

$$\begin{cases} \dfrac{\partial u}{\partial x}=3x^2y \\ \\ \dfrac{\partial u}{\partial y}=x^3+2y^4 \end{cases}$$

$$u(x, y)=\int 3x^2ydx=x^3y+\varphi (y)$$

$$\dfrac{\partial u}{\partial y}=x^3+\varphi' (y)=x^3+2y^4$$

$$\varphi' (y)=2y^4$$

The general solution of the exact differential equation is given by

$$x^3y+\dfrac{2}{5}y^5=C$$

where $C$ is an arbitrary constant.

4.

$$(\dfrac{1}{y})dx - (\dfrac{x}{y^2})dy = 0$$

$$\dfrac{\partial Q}{\partial x}=-\dfrac{1}{y^2}, \dfrac{\partial P}{\partial y}=-\dfrac{1}{y^2}$$

The given equation is exact because the partial derivatives are the same:

$$\dfrac{\partial Q}{\partial x}=-\dfrac{1}{y^2}=\dfrac{\partial P}{\partial y}$$

We have the following system of differential equations to find the function $u(x, y)$

$$\begin{cases} \dfrac{\partial u}{\partial x}=\dfrac{1}{y} \\ \\ \dfrac{\partial u}{\partial y}=-\dfrac{x}{y^2} \end{cases}$$

$$u(x, y)=\int (\dfrac{1}{y})dx=\dfrac{x}{y}+\varphi (y)$$

$$\dfrac{\partial u}{\partial y}=-\dfrac{x}{y^2}+\varphi' (y)=-\dfrac{x}{y^2}$$

$$\varphi' (y)=0$$

The general solution of the exact differential equation is given by

$$\dfrac{x}{y}=C$$

where $C$ is an arbitrary constant.

5.

$$(4x^3-y^3)dx - (3xy^2)dy = 0$$

$$\dfrac{\partial Q}{\partial x}=-3y^2, \dfrac{\partial P}{\partial y}=-3y^2$$

The given equation is exact because the partial derivatives are the same:

$$\dfrac{\partial Q}{\partial x}=-3y^2=\dfrac{\partial P}{\partial y}$$

We have the following system of differential equations to find the function $u(x, y)$

$$\begin{cases} \dfrac{\partial u}{\partial x}=4x^3-y^3 \\ \\ \dfrac{\partial u}{\partial y}=-3xy^2 \end{cases}$$

$$u(x, y)=\int (4x^3-y^3)dx=x^4-xy^3+\varphi (y)$$

$$\dfrac{\partial u}{\partial y}=-3xy^2+\varphi' (y)=-3xy^2$$

$$\varphi' (y)=0$$

The general solution of the exact differential equation is given by

$$x^4-xy^3=C$$

where $C$ is an arbitrary constant.