Homogeneous Differential Equations

Exercise 1.3

1. Find the general solution of the following differential equations.

(a) $y^{\prime} = \frac{xy}{x^{2}+y^{2}}$
(b) $y^{\prime} = \frac{xy}{(x+2y)^{2}}$
(c) $y^{\prime} = \frac{x^{2}+2xy-4y^{2}}{x^{2}-8xy-4y^{2}}$
(d) $(x^{2}-y^{2}e^{\frac{x}{y}})y^{\prime} = xy$
(e) $y^{\prime} = \frac{x+2y-1}{x+2y+7}$
(f) $y^{\prime} = \frac{x-y+8}{y-3x+2}$

2. Solve the following initial value problems.
(a) $(y - \sqrt{x^{2}+y^{2}})dx - xdy = 0, y(\sqrt{3}) = 1$
(b) $(y^{3}-x^{3})dx - xy^{2}dy = 0, y(1) = 2$
3. Example1.9 can not be solved by the method used in example1.8. Why?

Answer
1. (a) $xy, \ x^2 + y^2$ are homogeneous functions of the 2nd degree. So, divide by $x^2$.

$$y^{\prime} = \frac{xy}{x^{2}+y^{2}} = \frac{y/x}{1 + (y/x)^{2}}$$

Now let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies that

$$v^{\prime} x + v = \frac{v}{1 + v^2}$$

Rewrite this equation, we have

$$v^{\prime} x = \frac{v - v - v^3}{1 + v^2}$$

Separate the variables and integrate

$$\int \frac{1 + v^2}{v^3} dv = - \int \frac{1}{x} dx$$

and

$$\int (v^{-3} + v^{-1}) dv = - \log{x} + c$$

Thus,

$$\frac{v^{-2}}{2} + \log{\vert v\vert} = - \log{x} + c$$

Now substitute $v = y/x$, we have

$$\frac{1}{2}(\frac{x}{y})^{2} + \log{\vert\frac{y}{x}\vert} + \log{x} = c$$

Simplify the equation, we obtain

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

(b) $xy, \ (x + y)^{2}$ are homogeneous functions of the 2nd degree. So, divide by $x^2$.

$$y^{\prime} = \frac{xy}{(x + y)^{2}} = \frac{y/x}{(1 + (2y/x))^{2}}.$$

Now let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies that

$$v^{\prime} x + v = \frac{v}{(1 + 2v)^{2}}.$$

Rewrite this equation, we have

$$v^{\prime} x = \frac{v - v - 4v^2 - 4v^3}{1 + v^2}$$

Thus,

$$\int \frac{1 + 4v + 4v^2}{4v^3 + 4v^2}dv = - \int \frac{1}{x} dx.$$

Here, we use the partial fraction on the left-hand side.

$$\frac{1 + 4v + 4v^2}{4v^3 + 4v^2} = \frac{4v^2 + 4v + 1}{4v^2(v + 1)} = \frac{1}{4}(\frac{Av + B}{v^2} + \frac{C}{v+1}).$$

Thus, $A = 3, B=1, C=1$. Therefore,

$$\int \frac{1}{4}(\frac{3v + 1}{v^2} + \frac{1}{1+v}) dv = - \int \frac{1}{x} dx$$

Evaluate the integrals, we have

$$\frac{1}{4}(3\log{v} - \frac{1}{v} + \log{\vert 1 + v\vert}) = - \log{x} + c$$

Multiply both sides by 4.

$$\log{{v^3}(1 + v)} - \frac{1}{v} = - \log{x^{4}} + C$$

Substitute $v = y/x$. Then

$$\log{[(\frac{y}{x})^{3} + (\frac{y}{x})^{4}]} - \frac{x}{y} = - \log{x^{4}} + C$$

Therefore,

$$\log{(\frac{y^{3}}{x^{3}} + \frac{y^4}{x^{4}}) x^{4}} - \frac{x}{y} = C$$

Simplify to obtain.

$$\log{(xy^3 + y^4)} - \frac{x}{y} = C \ \ \$$

(c) $x^2 + 2xy - 4y^2, \ x^2 - 8xy - 4y^2$ are homogeneous functions of the 2nd degree. So, divide by $x^2$.

$$y^{\prime} = \frac{1 + 2y/x - 4y^{2}/x^{2}}{1 - 8y/x - 4y^{2}/x^{2}}.$$

Let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies that

$$v^{\prime} x + v = \frac{1 + 2v - 4v^2}{1 - 8v - 4v^2}.$$

Rewrite this equation, we have

$$v^{\prime} x = \frac{1 + 2v - 4v^2 - v + 8v^2 + 4v^3}{1 - 8v - 4v^2}$$

Integrate both sides.

$$\int \frac{4v^2 + 8v - 1}{4v^3 + 4v^2 + v + 1} dv = - \int \frac{1}{x} dx$$

or

$$\int (\frac{-1}{v + 1} + \frac{8v}{4v^2 + 1}) dv = - \int \frac{1}{x} dx$$

Evaluate the integrals.

$$- \log{\vert v + 1\vert} + \log{\vert 4v^2 + 1\vert} = - \log{\vert x\vert} + c$$

Here note that a difference of the logarithm is the logarithm of a quotients.

$$\log{\vert\frac{4v^2 +1}{v + 1}\vert} + \log{\vert x\vert} = c$$

Also, a sum of the logarithms is the logarithm of a product.

$$\log{\vert\frac{4xv^2 + x}{v + 1}\vert} = c$$

Substitute $v = y/x$.

$$\log{\vert\frac{4x(y/x)^2 + x}{(y/x) + 1}\vert} = c$$

or

$$\log{\vert\frac{4y^2 + x^2}{y + x}\vert} = c$$

Thus, we have

$$x^2 + 4y^2 = C(x + y) \ \ \$$

(d) $x^2 - y^{2}e^{x/y}, \ xy$ are homogeneous functions of the 2nd degree.

$$(1 - (\frac{y}{x})^{2}e^{\frac{y}{x}})y^{\prime} = \frac{y}{x}.$$

Let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies that

$$v^{\prime} x + v = \frac{v}{1 - v^{2}e^{\frac{1}{v}}}.$$

Rewrite this equation,

$$v^{\prime} x = \frac{v - v + v^{3}e^{\frac{1}{v}}}{1 - v^{2}e^{\frac{1}{v}}}$$

Integrate

$$\int \frac{1 - v^{2}e^{\frac{1}{v}}}{v^{3}e^{\frac{1}{v}}} dv = \int \frac{1}{x} dx$$

or

$$\int \frac{1}{v^{3}}e^{-\frac{1}{v}} dv - \int \frac{1}{v} dv = \int \frac{1}{x} dx$$

Using the integration by parts, we have

$$\left(\begin{array}{ll} u = \frac{1}{v} & dw = \frac{1}{v^{2}}e^... ...}{v}}dv\\ du = -\frac{dv}{v^{2}} & w = e^{-\frac{1}{v}} \end{array} \right)$$

$$\frac{1}{v}e^{-\frac{1}{v}} + \int \frac{1}{v^{2}}e^{-\frac{1}{v}}dv - \log{\vert v\vert} = \log{\vert x\vert} + c$$

Then,

$$\frac{1}{v}e^{-\frac{1}{v}} + e^{-\frac{1}{v}} - \log{\vert v\vert} = \log{\vert x\vert} + c$$

Substitute $v = y/x$.

$$\frac{x}{y}e^{-\frac{x}{y}} + e^{-\frac{x}{y}} - \log{\frac{y}{x}} - \log{\vert x\vert} = C$$

Therefore,

$$e^{-\frac{x}{y}}(\frac{x}{y} + 1) - \log{y} = C \ \ \$$

(e) $x + 2y - 1$ and $x + 2y + 7$ are not homogeneous functions，But they contain $x + 2y$. So, we let $u = x + 2y$. Then $u^{\prime} = 1 + 2y^{\prime}$ and

$$\frac{u^{\prime} - 1}{2} = \frac{u - 1}{u + 7}.$$

Now rewrite this equation.

$$u^{\prime} = \frac{2u - 2 + u+ 7}{u + 7} = \frac{3u + 5}{u + 7}$$

Separate the variables.

$$\int \frac{u + 7}{3u + 5} du = \int dx$$

Evaluate the integrals.

$$\frac{1}{3}\int(1 + \frac{16}{3u + 5}) du = \int dx$$

Thus,

$$\frac{1}{3}u + \frac{16}{9}\log{\vert 3u + 5\vert} = x + c$$

Substitute $u = x + 2y$. Then

$$\frac{1}{3}(x + 2y) + \frac{16}{9}\log{\vert 3(x + 2y) + 5\vert} = x + c$$

Therefore,

$$2x - 2y - \frac{16}{3}\log{\vert 3x + 6y +5\vert} = C \ \ \$$

(f) $x - y + 8$ and $y - 3x + 2$ are not homogeneous functions. But once the constant term is gone, it becomes homogeneous function. So, we solve the following system of linear equations:

$$\left\{\begin{array}{ll} x - y + 8 & = 0 \\ -3x + y + 2 & = 0 \end{array}\right.$$

implies that

$$x = \frac{\left \vert \begin{array}{cc} -8 & -1\\ -2 & 1 \end... ...eft \vert \begin{array}{cc} 1 & -1\\ -3 & 1 \end{array} \right \vert} = 13$$

Thus, we use the change of coordinates so that $x = 5, y = 13$ becomes the origin. Let $x = X + 5, y = Y + 13$. Then we have

$$\frac{dY}{dX} = \frac{dY}{dx}\frac{dx}{dX} = \frac{dy}{dx} = \frac{X - Y}{Y - 3X}$$

This is homogeneous differential equation. So, let $v = Y/X$ and $Y = vX, \ Y^{\prime} = v^{\prime}X + v$ implies that

$$v^{\prime}X + v = \frac{1 - v}{v - 3}.$$

Rewrite this equation,

$$v^{\prime}X = \frac{1 - v - v^{2} + 3v}{v - 3}$$

Integrate

$$\int \frac{v - 3}{v^2 - 2v -1} dv = - \int \frac{1}{X} dX$$

$$\frac{1}{2} \int (\frac{1 + \sqrt{2}}{v - 1 + \sqrt{2}} + \frac{1 - \sqrt{2}}{v - 1 - \sqrt{2}}) dv = - \log{\vert X\vert} + c$$

Evaluate the integrals,

$$\frac{1}{2}((1 + \sqrt{2})\log{\vert v - 1 + \sqrt{2}\vert} + (1 - \sqrt{2})\log{\vert v - 1 - \sqrt{2}\vert}) = - \log{\vert X\vert} + c$$

Substitute $v = Y/X$, we have

$$\frac{1}{2}((1 + \sqrt{2})\log{\vert \frac{Y}{X} - 1 + \sqrt{2}\v... ...sqrt{2})\log{\vert\frac{Y}{X} - 1 - \sqrt{2}\vert}) = - \log{\vert X\vert} + c$$

Now replace $X = x - 5, Y = y - 13$. Then

$$(1 + \sqrt{2})\log{\vert\frac{y - 13}{x - 5} - 1 + \sqrt{2}\vert}... ...t\frac{y - 13}{x - 5} - 1 - \sqrt{2}\vert} + 2\log{\vert x - 5\vert} = C \ \ \$$

2. (a) $y - \sqrt{x^2 + y^2}, x$ are homogeneous functions of 1st degree. So, divide by $x$.

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

Now let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies that

$$v^{\prime}x + v = v - \sqrt{1 + v^2}.$$

Rewrite this equation.

$$v^{\prime}x = - \sqrt{1 + v^2}$$

Integrate both sides.

$$\int \frac{dv}{\sqrt{1 + v^2}} = - \int \frac{1}{x} dx$$

Evaluate the integrals.

$$\log{\vert v + \sqrt{1 + v^2}\vert} = - \log{\vert x\vert} + c$$

Substitute $v = y/x$.

$$\log{\vert \frac{y}{x} + \sqrt{1 + (\frac{y}{x})^2}\vert} + \log{\vert x\vert} = c$$

or

$$\log{(y + \sqrt{x^2 + y^2})} = C$$

Take the exponential.

$$y + \sqrt{x^2 + y^2} = C$$

Using the initial condition $y(\sqrt{3}) = 1$, we have $1 + \sqrt{4} = C$. Thus,

$$y + \sqrt{x^2 + y^2} = 3 \ \ \$$

(b) $y^3 - x^3, xy^2$ are homogeneous functions of the 3rd degree. So, divide by $x^3$.

$$((\frac{y}{x})^{3} - 1) dx - \frac{y^2}{x^2} dy = 0.$$

Let $v = y/x$. Then $y = vx, \ y^{\prime} = v^{\prime} x + v$ implies

$$v^{\prime}x + v = \frac{v^3 - 1}{v^2} = v - \frac{1}{v^2}.$$

Rewrite this equation.

$$v^{\prime}x = - \frac{1}{v^2}$$

Integrate both sides.

$$\int v^2 dv = - \int \frac{1}{x} dx$$

Evaluate the integrals.

$$\frac{v^3}{3} = - \log{\vert x\vert} + c$$

Substitute $v = y/x$.

$$(\frac{y}{x})^{3} = - 3\log{\vert x\vert} + C$$

or

$$(\frac{y}{x})^{3} + 3\log{x} = C$$

Using the initial condition $y(1) = 2$, we have $8 = C$. Therefore,

$$(\frac{y}{x})^{3} + 3\log{x} = 8 \ \ \$$

3. In example1.8, we are able to get rid of the constant term by moving the coordinates. But in example1.9, two equations $x + y - 1 = 0, \ x + y + 1 = 0$ represents the parallel lines. So, we can not find the intersection.