# Differential Equations

## First Order Equations # Homogeneous Equations

• ### Definition of Homogeneous Differential Equation

A first order differential equation

$\frac{{dy}}{{dx}} = f\left( {x,y} \right)$

is called homogeneous equation, if the right side satisfies the condition

$f\left( {tx,ty} \right) = f\left( {x,y} \right)$

for all $$t.$$ In other words, the right side is a homogeneous function (with respect to the variables $$x$$ and $$y$$) of the zero order:

${f\left( {tx,ty} \right) }={ {t^0}f\left( {x,y} \right) }={ f\left( {x,y} \right).}$

A homogeneous differential equation can be also written in the form

$y’ = f\left( {\frac{x}{y}} \right),$

or alternatively, in the differential form:

${P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy }={ 0,}$

where $$P\left( {x,y} \right)$$ and $$Q\left( {x,y} \right)$$ are homogeneous functions of the same degree.

### Definition of Homogeneous Function

A function $$P\left( {x,y} \right)$$ is called a homogeneous function of the degree $$n$$ if the following relationship is valid for all $$t \gt 0:$$

$P\left( {tx,ty} \right) = {t^n}P\left( {x,y} \right).$

### Solving Homogeneous Differential Equations

A homogeneous equation can be solved by substitution $$y = ux,$$ which leads to a separable differential equation.

A differential equation of kind

${\left( {{a_1}x + {b_1}y + {c_1}} \right)dx }+{ \left( {{a_2}x + {b_2}y + {c_2}} \right)dy} ={ 0}$

is converted into a separable equation by moving the origin of the coordinate system to the point of intersection of the given straight lines. If these straight lines are parallel, the differential equation is transformed into separable equation by using the change of variable:

$z = ax + by.$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Solve the differential equation $$\left( {2x + y} \right)dx$$ $$- xdy = 0.$$

### Example 2

Solve the differential equation $$xy’ = y\ln {\large\frac{y}{x}\normalsize}.$$

### Example 3

Solve the differential equation $$\left( {xy + {y^2}} \right)y’$$ $$= {y^2}.$$

### Example 4

Solve the differential equation $$y’ = {\large\frac{y}{x}\normalsize} – {\large\frac{x}{y}\normalsize}.$$

### Example 5

Find the general solution of the differential equation $$\left( {{x^3} + x{y^2}} \right)y’$$ $$= {y^3}.$$

### Example 6

Solve the equation $$y’ = {\large\frac{{2x + 1}}{{3y + x + 2}}\normalsize}.$$

### Example 7

Find the general solution of the differential equation $$y’ = {\large\frac{{x – y + 3}}{{x – y}}\normalsize}.$$

### Example 1.

Solve the differential equation $$\left( {2x + y} \right)dx$$ $$- xdy = 0.$$

Solution.

It is easy to see that the polynomials $$P\left( {x,y} \right)$$ and $$Q\left( {x,y} \right),$$ respectively, at $$dx$$ and $$dy,$$ are homogeneous functions of the first order. Therefore, the original differential equation is also homogeneous.

Suppose that $$y = ux,$$ where $$u$$ is a new function depending on $$x.$$ Then

${dy = d\left( {ux} \right) }={ udx + xdu.}$

Substituting this into the differential equation, we obtain

${\left( {2x + ux} \right)dx }-{ x\left( {udx + xdu} \right) }={ 0.}$

Hence,

$\require{cancel} {2xdx + \cancel{uxdx} – \cancel{xudx} }-{ {x^2}du = 0.}$

Dividing both sides by $$x$$ yields:

${xdu = 2dx\;\;\text{or}\;\;}\kern-0.3pt{du = 2\frac{{dx}}{x}.}$

When dividing by $$x,$$ we could lose the solution $$x = 0.$$ The direct substitution shows that $$x = 0$$ is indeed a solution of the given differential equation.

Integrate the latter expression to obtain:

${\int {du} = 2\int {\frac{{dx}}{x}} \;\;\text{or}\;\;}\kern-0.3pt{u = 2\ln \left| x \right| + C,}$

where $$C$$ is a constant of integration.

Returning to the old variable $$y,$$ we can write:

${y = ux }={ x\left( {2\ln \left| x \right| + C} \right).}$

Thus, the equation has two solutions:

${y = x\left( {2\ln \left| x \right| + C} \right),\;\;}\kern-0.3pt{x = 0.}$

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