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

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,

\[2xdx + \cancel{uxdx} - \cancel{xudx} - {x^2}du = 0.\]

Dividing both sides by \(x\) yields:

\[xdu = 2dx\;\;\text{or}\;\; du = 2\frac{{dx}}{x}.\]

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:

\[\int {du} = 2\int {\frac{{dx}}{x}} \;\;\text{or}\;\; u = 2\ln \left| x \right| + C,\]

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:

\[y = x\left( {2\ln \left| x \right| + C} \right),\;\; x = 0.\]

Example 2.

Solve the differential equation \[xy^\prime = y\ln {\frac{y}{x}}.\]

Solution.

We notice that the root \(x = 0\) does not belong to the domain of the differential equation. Rewrite the equation in the form:

\[y' = \frac{y}{x}\ln \frac{y}{x} = f\left( {\frac{y}{x}} \right).\]

As you can see, this equation is homogeneous.

Make the substitution \(y = ux.\) Hence,

\[y' = \left( {ux} \right)^\prime = u'x + u.\]

Substituting this expression into the equation gives:

\[x\left( {u'x + u} \right) = ux\ln \frac{{ux}}{x}.\]

Divide by \(x \ne 0\) to get:

\[u'x + u = u\ln u,\;\; \Rightarrow \frac{{du}}{{dx}}x = u\ln u - u,\;\; \Rightarrow \frac{{du}}{{dx}}x = u\left( {\ln u - 1} \right).\]

We obtain the separable equation:

\[\frac{{du}}{{u\left( {\ln u - 1} \right)}} = \frac{{dx}}{x}.\]

The next step is to integrate the left and the right side of the equation:

\[\int {\frac{{du}}{{u\left( {\ln u - 1} \right)}}} = \int {\frac{{dx}}{x}} ,\;\; \Rightarrow \int {\frac{{d\left( {\ln u} \right)}}{{\ln u - 1}}} = \int {\frac{{dx}}{x}} ,\;\; \Rightarrow \int {\frac{{d\left( {\ln u - 1} \right)}}{{\ln u - 1}}} = \int {\frac{{dx}}{x}} .\]

Hence,

\[\ln\left| {\ln u - 1} \right| = \ln \left| x \right| + C.\]

Here the constant \(C\) can be written as \(\ln {C_1}\) \(\left( {{C_1} \gt 0} \right).\) Then

\[\ln\left| {\ln u - 1} \right| = \ln \left| x \right| + \ln {C_1},\;\; \Rightarrow \ln\left| {\ln u - 1} \right| = \ln \left| {{C_1}x} \right|,\;\; \Rightarrow \ln u - 1 = \pm {C_1}x,\;\; \Rightarrow \ln u = 1 \pm {C_1}x\;\; \text{or}\;\;u = {e^{1 \pm {C_1}x}}.\]

Thus, we have got two solutions:

\[u = {e^{1 + {C_1}x}}\;\; \text{and}\;\;u = {e^{1 - {C_1}x}}.\]

If \({C_1} = 0,\) the answer is \(y = xe\) and we can make sure that it is also a solution to the equation. Indeed, substituting

\[y = xe,\;\;y' = e\]

into the differential equation, we obtain:

\[xe = xe\ln \frac{{\cancel{x}e}}{\cancel{x}},\;\; \Rightarrow xe = xe\ln e,\;\; \Rightarrow xe = xe.\]

Then all the solutions can be represented by one formula:

\[y = x{e^{1 + Cx}},\]

where \(C\) is an arbitrary real number.

See more problems on Page 2.