Answer to Question #92510 in Differential Equations for MWALE BORNFACE

Question

Kindly assist me solving the following questions
1. (x^2y^2+xy)ydx + (x^2y^2-1)xdy=0
2. (x^2+y^2)dx + xdy=0
3. dy/dx +y=3y^3e^x

Answer 1

1.

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

$$x=0$$ $$y=0$$ $$xy+1=0=>y=-{1 \over x}$$

Or

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

Using an Integrating Factor

$$P(x,y)=y^2, Q(x,y)=xy-1$$

$${\eth P \over \eth y}=2y, {\eth Q \over \eth x}=y, {\eth P \over \eth y}\not={\eth Q \over \eth x}$$ $${\eth P \over \eth y}-{\eth Q \over \eth x}=2y-y=y$$

Integrating Factor

$$\mu=\mu(y)$$

$${1 \over \mu}{d\mu \over dy}=-{1 \over P}({\eth P \over \eth y}-{\eth Q \over \eth x})$$

$${1 \over \mu}{d\mu \over dy}=-{1 \over y^2}(2y-y)$$

$${d\mu \over \mu}=-{dy \over y}$$

$$\int {d\mu \over \mu}=\int (-{dy \over y})=>\ln|\mu|=\ln|{1 \over y}|+\ln C$$

We choose

$$\mu={1 \over y}$$

Multiplying the original differential equation by $\mu=1/y$  produces the exact equation:

$$ydx+(x-{1 \over y})dy=0$$

Indeed, now we have

$${\eth P \over \eth y}=1, {\eth Q \over \eth x}=1, {\eth P \over \eth y}={\eth Q \over \eth x}$$

Solve the resulting equation. The function $u(x,y)$ can be found from the system of equations:

$$\begin{cases} {\eth u \over \eth x}=y \\ {\eth u \over \eth y}=x-{1 \over y} \end{cases}$$

It follows from the first equation that

$$u=\int y dx=xy+\varphi (y)$$

Substitute this in the second equation to determine $\varphi:$

$${\eth u \over \eth y}=x+ {d \varphi \over d y}=x-{1 \over y}$$

$$\int d\varphi=\int (-{1 \over y})dy=>\varphi (y)=\ln({1 \over |y|})$$

Thus, the original differential equation has the following solutions:

$$x=0$$ $$y=0$$ $$y=-{1 \over x}$$ $$xy+\ln({1 \over |y|})=C_1$$

where $C_1$ is a constant.

2.

$$(x^2+y^2-y)dx+xdy=0$$

We can make sure that this equation is not exact:

$$P(x,y)=x^2+y^2-y, Q(x,y)=x$$

$${\eth P \over \eth y}=2y-1, {\eth Q \over \eth x}=1, {\eth P \over \eth y}\not={\eth Q \over \eth x}$$

The difference of the partial derivatives is

$${\eth P \over \eth y}-{\eth Q \over \eth x}=2y-1-1=2y-2$$

Using the integrating factor $\mu=x^2+y^2$ we find that

$${\eth z \over \eth y}=2y, {\eth z \over \eth x}=2x$$

Calculate the following expression:

$$Q{\eth z \over \eth x}-P {\eth z \over \eth y}=x(2x)-(x^2+y^2-y)(2y)=$$ $$=2x^2+2y^2-2y(x^2+y^2)=(x^2+y^2)(2-2y)$$

As a result, we obtain the differential equation for the function $\mu(z):$

$${1 \over \mu}{d\mu \over dz}={1 \over (Q{\eth z \over \eth x}-P {\eth z \over \eth y})}({\eth P \over \eth y}-{\eth Q \over \eth x})=$$

$$={1 \over (x^2+y^2)(2-2y)}(2y-2)=-{1 \over {x^2+y^2}}=-{1 \over z}$$

By integrating we get the function $\mu(z):$

$$\int {d\mu \over \mu}=\int (-{dz \over z})=>\ln|\mu|=-\ln|z|+\ln C$$

We choose

$$\mu={1 \over z}$$

We can choose the integrating factor

$$\mu=-{1 \over z}=-{1 \over {x^2+y^2}}$$

$$-{1 \over {x^2+y^2}}(x^2+y^2-y)dx-{1 \over {x^2+y^2}}xdy=0$$

Or

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

The general solution $u(x,y)=C$ is defined by the following system of equations:

$$\begin{cases} {\eth u \over \eth x}=1-{y\over {x^2+y^2}} \\ {\eth u \over \eth y}={x \over {x^2+y^2}} \end{cases}$$

Integrating the first equation with respect to x produces:

$$u(x,y)=\int{(1-{y\over {x^2+y^2}})}dx=x-y\int{{1\over {x^2+y^2}}}dx=$$

$$=x-\arctan({x \over y})+\varphi(y)$$

Substituting this in the second equation, we have:

$${\eth u \over \eth y}={x\over {x^2+y^2}}+ {d \varphi \over d y}={x\over {x^2+y^2}}$$

$${d \varphi \over d y}=0$$

Thus, the general solution of differential equation in implicit form is defined by the formula:

$$x-\arctan({x \over y})=C$$

where $C$ is a constant.

3.

$${dy \over dx}+y=3y^3e^x$$ $$y=0$$

Or

$$y^{-3}{dy \over dx}+y^{-2}=3e^x$$

The associated homogeneous differential equation is

$$y^{-3}{dy \over dx}+y^{-2}=0$$

$${dy \over dx}+y=0$$

$${dy \over y}=-dx$$ $$y=C_1e^{-x}$$

$${dy \over dx}={dC_1 \over dx}e^{-x}-C_1e^{-x}$$

Substitute

$${dC_1 \over dx}e^{-x}-C_1e^{-x}+C_1e^{-x}=3C_1^3e^{-3x}e^x$$

$${dC_1 \over dx}=3C_1^3e^{-x}$$

$${dC_1 \over C_1^3}=3e^{-x}dx$$

$$\int {dC_1 \over C_1^3}=\int3e^{-x}dx$$

$$-{1 \over2 C_1^2}=-3e^{-x}dx-{C_2 \over2}$$

$$C_1={1\over \sqrt{6e^{-x}dx+C_2}}$$

Thus, the original differential equation has the following solutions:

$$y=0$$ $$y={1\over e^x \sqrt{6e^{-x}dx+C_2}}$$

where $C_1$ is a constant.