Answer to Question #214053 in Differential Equations for jen

Question

Find the general solutions of the following equations.

LINEAR DE

1. dx/dy - 4x/y = y⁵
2. dy/dx + y/x = cos x + (sinx/x)
3. dx/dy + x(2-y)/y = 1/y

BERNOULLI'S EQUATION

1. dy/dx - 3y/2x = y³/2x³
2. dy/dx - y/2x = -x²/y
3. dy/dx+ y = 2xy²e^x

Answer 1

Solution.

All equations will be done by the method of variation of a constant.

1.1

$$\frac{dx}{dy}-\frac{4x}{y}=y^5$$

1) $\frac{dx}{dy}-\frac{4x}{y}=0$

$\frac{dx}{x}=\frac{4dy}{y}$

$\int\frac{dx}{x}=\int\frac{4dy}{y}$

$x=Cy^4,$

where C is some constant.

2) Let be $x=C(y)y^4,$

then$x'=C'(y)y^4+4C(y)y^3.$

We will have

$C'y^4+4Cy^3-\frac{4}{y}Cy^4=y^5.$

From here $C'(y)=y.$

So, $C(y)=\frac{y^2}{2}.$

And, $x=\frac{y^2}{2}y^4=\frac{y^6}{2}.$

3)

$$x=Cy^4+\frac{y^6}{2}.$$

1.2

$$\frac{dy}{dx}+\frac{y}{x}=\cos{x}+\frac{\sin{x}}{x}$$

1) $\frac{dy}{dx}+\frac{y}{x}=0$

$\frac{dy}{y}=-\frac{dx}{x}$

$\int\frac{dy}{y}=-\int\frac{dx}{x}$

$yx=C$

$y=\frac{C}{x},$

where C is some constant.

2) Let be $y=\frac{C(x)}{x},$

then $y'=\frac{C'(x)x-C(x)}{x^2}.$

We will have

$\frac{C'x-C}{x^2}+\frac{C}{x^2}=\cos{x}+\frac{\sin{x}}{x}$

From here $C'(x)=x\cos{x}+\sin{x}.$

So, $C(x)=x\sin{x}.$

And, $y=\frac{x\sin{x}}{x}=\sin{x}.$

3)

$$y=\frac{C}{x}+\sin{x}.$$

1.3

$$\frac{dx}{dy}+\frac{x(2-y)}{y}=\frac{1}{y}$$

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

$\frac{dx}{x}=-\frac{2-y}{y}dy$

$\int\frac{dx}{x}=-\int\frac{2-y}{y}dy$

$x=\frac{Ce^y}{y^2},$

where C is some constant.

2) Let be $x=\frac{C(y)e^y}{y^2},$

then$x'=\frac{(C'(y)e^y+C(y)e^y)y^2-C(y)e^y2y}{y^4}.$

We will have

$\frac{(C'(y)e^y+C(y)e^y)y^2-C(y)e^y2y}{y^4}+\frac{C(y)e^y}{y^2}\frac{2-y}{y}=\frac{1}{y}.$

From here $C'(y)=\frac{y}{e^y}.$

So, $C(y)=-\frac{y+1}{e^y}.$

And, $x=-\frac{y+1}{y^2}.$

3)

$$x=C\frac{e^y}{y^2}-\frac{y+1}{y^2}.$$ $$2.1. \frac{dy}{dx}-\frac{3y}{2x}=\frac{y^3}{2x^3}$$

Divide both parts of equation by $y^3.$

$$\frac{y'}{y^3}-\frac{3}{2xy^2}=\frac{1}{2x^3}$$

Substitution $z=\frac{1}{y^2}.$

Then $z'+\frac{3z}{x}=-\frac{1}{x^3}.$

1) $\frac{dz}{dx}+\frac{3z}{x}=0$

$\frac{dz}{dx}=-\frac{3z}{x}$

$\int\frac{dz}{dx}=-\int\frac{3z}{x}$

$z=\frac{C}{x^3},$

where C is some constant.

2) Let be $z=\frac{C(x)}{x^3},$

then $z'=\frac{C'(x)x^3-3x^2C(x)}{x^6}.$

We will have

$\frac{C'}{x^3}-\frac{3C}{x^4}+\frac{3C}{x^4}=-\frac{1}{x^3}$

From here $C'(x)=-1.$

So, $C(x)=-x.$

And, $z=-\frac{1}{x^2}.$

3)

$$z=\frac{C}{x^3}-\frac{1}{x^2}.$$

From here

$$y=\pm \sqrt{\frac{x^3}{C-x}}.$$

2.2

$$\frac{dy}{dx}-\frac{y}{2x}=-\frac{x^2}{y}$$

Divide both parts of equation by $y^{-1}$ .

$$\frac{y'}{y^{-1}}-\frac{y^2}{2x}=-x^2$$

Substitution $z={y^2}.$

Then $z'-\frac{z}{x}=-2{x^2}.$

1) $\frac{dz}{dx}-\frac{z}{x}=0$

$\frac{dz}{z}=\frac{dx}{x}$

$\int\frac{dz}{dx}=-\int\frac{3z}{x}$

$z={C}{x},$

where C is some constant.

2) Let be $z={C(x)}{x},$

then $z'=C'(x)x+C(x).$

We will have

${C'}{x}+C-\frac{Cx}{x}=-2x^2.$

From here $C'(x)=-2x.$

So, $C(x)=-x^2.$

And, $z=-{x^3}.$

3)

$$z={C}{x}-{x^3}.$$

From here

$$y=\pm \sqrt{{Cx}-{x^3}}.$$

2.3

$$\frac{dy}{dx}+y=2xy^2e^x$$

Divide both parts of equation by $y^2.$

$$\frac{y'}{y^2}+\frac{1}{y}={2xe^x}$$

Substitution $z=\frac{1}{y}.$

Then $z'-{z}=-2xe^x.$

1) $\frac{dz}{dx}-z=0$

$\frac{dz}{z}=dx$

$\int\frac{dz}{z}=\int{dx}$

$z={C}{e^x},$

where C is some constant.

2) Let be $z={C(x)}{e^x},$

then $z'={C'(x)e^x+C(x)}{e^x}.$

We will have

${C'}{e^x}+{C}{e^x}-{C}{e^x}=-2xe^x$

From here $C'(x)=-2x.$

So, $C(x)=-x^2.$

And, $z=-{x^2}e^x.$

3)

$${C}{e^x}-{x^2}e^x.$$

From here

$$y= \frac{e^{-x}}{C-x^2}.$$