Homogeneous Differential Equations

A differential equation can be homogeneous in either of two respects.

A first order differential equation is said to be homogeneous if it may be written

$f(x,y)\,dy=g(x,y)\,dx,$

where f and g are homogeneous functions of the same degree of x and y. In this case, the change of variable y = ux leads to an equation of the form

${\frac {dx}{x}}=h(u)\,du,$

which is easy to solve by integration of the two members.

Otherwise, a differential equation is homogeneous if it is a homogeneous function of the unknown function and its derivatives. In the case of linear differential equations, this means that there are no constant terms. The solutions of any linear ordinary differential equation of any order may be deduced by integration from the solution of the homogeneous equation obtained by removing the constant term.

Homogeneous first-order differential equations

A first-order ordinary differential equation in the form:

$M(x,y)\,dx+N(x,y)\,dy=0$

is a homogeneous type if both functions M(x, y) and N(x, y) are homogeneous functions of the same degree n. That is, multiplying each variable by a parameter λ, we find

$M(\lambda x,\lambda y)=\lambda ^{n}M(x,y)\quad {\text{and}}\quad N(\lambda x,\lambda y)=\lambda ^{n}N(x,y)\,.$

Thus,

${\frac {M(\lambda x,\lambda y)}{N(\lambda x,\lambda y)}}={\frac {M(x,y)}{N(x,y)}}\,.$

Solution method

In the quotient ${\textstyle {\frac {M(tx,ty)}{N(tx,ty)}}={\frac {M(x,y)}{N(x,y)}}}$, we can let t = Template:Sfrac to simplify this quotient to a function f of the single variable Template:Sfrac:

${\frac {M(x,y)}{N(x,y)}}={\frac {M(tx,ty)}{N(tx,ty)}}={\frac {M(1,y/x)}{N(1,y/x)}}=f(y/x)\,.$

That is

${\frac {dy}{dx}}=-f(y/x).$

Introduce the change of variables y = ux; differentiate using the product rule:

${\frac {dy}{dx}}={\frac {d(ux)}{dx}}=x{\frac {du}{dx}}+u{\frac {dx}{dx}}=x{\frac {du}{dx}}+u.$

This transforms the original differential equation into the separable form

$x{\frac {du}{dx}}=-f(u)-u,$

or

${\frac {1}{x}}{\frac {dx}{du}}={\frac {-1}{f(u)+u}},$

which can now be integrated directly: ln x equals the antiderivative of the right-hand side (see ordinary differential equation).

Special case

A first order differential equation of the form (a, b, c, e, f, g are all constants)

$\left(ax+by+c\right)dx+\left(ex+fy+g\right)dy=0$

where af ≠ be can be transformed into a homogeneous type by a linear transformation of both variables (α and β are constants):

$t=x+\alpha ;\;\;z=y+\beta \,.$

Homogeneous linear differential equations

Template:See also A linear differential equation is homogeneous if it is a homogeneous linear equation in the unknown function and its derivatives. It follows that, if φ(x) is a solution, so is cφ(x), for any (non-zero) constant c. In order for this condition to hold, each nonzero term of the linear differential equation must depend on the unknown function or any derivative of it. A linear differential equation that fails this condition is called inhomogeneous.

A linear differential equation can be represented as a linear operator acting on y(x) where x is usually the independent variable and y is the dependent variable. Therefore, the general form of a linear homogeneous differential equation is

$L(y)=0$

where L is a differential operator, a sum of derivatives (defining the "0th derivative" as the original, non-differentiated function), each multiplied by a function f_i of x:

$L=\sum _{i=0}^{n}f_{i}(x){\frac {d^{i}}{dx^{i}}}\,,$

where f_i may be constants, but not all f_i may be zero.

For example, the following linear differential equation is homogeneous:

$\sin(x){\frac {d^{2}y}{dx^{2}}}+4{\frac {dy}{dx}}+y=0\,,$

whereas the following two are inhomogeneous:

$2x^{2}{\frac {d^{2}y}{dx^{2}}}+4x{\frac {dy}{dx}}+y=\cos(x)\,;$

$2x^{2}{\frac {d^{2}y}{dx^{2}}}-3x{\frac {dy}{dx}}+y=2\,.$

The existence of a constant term is a sufficient condition for an equation to be inhomogeneous, as in the above example.

Characteristic equation

If the equation is linear homogeneous and further $p(t),q(t)$ are constant, then the equation is referred to as a constant-coefficients equation:

$$ay''+by'+cy=0$$

and we can apply the method of characteristic equations to solve such an equation. Note that $a$ is assumed to be non-zero since we are working with a second order equation.

Method formal steps

1. We assume that the solution is of the form $y(t)=e^{rt}$ (this is called making an ansatz). This gives
   $$(ar^{2}+br+c)e^{rt}=0\Longrightarrow \,ar^{2}+br+c=0,$$
   which equation is called the characteristic equation.
2. So to solve the above ODE, it suffices to find the two roots $r_{1},r_{2}$.
3. Then the general solution is of the form:
   $$y(t)=c_{1}e^{r_{1}t}+c_{2}e^{r_{2}t}.$$

Example-presenting the method

Consider a mass $m$ hanging at rest on the end of a vertical spring of length $l$, spring constant $k$ and damping constant $\gamma$. Let $u\left(t\right)$ denote the displacement, in units of feet, from the equilibrium position. Note that since $u(t)$ represents the amount of displacement from the spring's equilibrium position (the position obtained when the downward force of gravity is matched by the will of the spring to not allow the mass to stretch the spring further) then $u(t)$ should increase downward. Then by Newton's Third Law one can obtain the equation

$$mu''(t)+\gamma u'(t)+ku(t)=F(t),$$

where $F\left(t\right)$ is any external force, which for simplicity we will assume to be zero.

1. First we obtain the characteristic equation:
   $$mr^{2}+\gamma r+k=0.$$
2. Suppose that $m=1{\text{lb}},\gamma =5{\text{lb}}/{\text{ft}}/{\text{s}}$ and $k=6{\text{lb}}/{\text{ft}}$ then we obtain the roots $r_{1}=-2$, $r_{2}=-3$.
3. Therefore, the general solution will be
   $$u(t)=c_{1}e^{-2t}+c_{2}e^{-3t}.$$
4. Further if $u(0)=0,\,u'(0)=1$ we obtain $c_{1}=1,c_{2}=-1$:
   $$u(t)=e^{-2t}-e^{-3t}.$$

Examples

• Consider the IVP

$$4y''-y=0,\,\,y(-2)=1,\,\,y'(-2)=-1.$$

1. We obtain the characteristic equation $4r^{2}-1=0\Rightarrow r=\pm {\frac {1}{2}}$ and so the general solution will be
   $$y(t)=c_{1}e^{\frac {t}{2}}+c_{2}e^{-{\frac {t}{2}}}.$$
2. Using the initial conditions we obtain:
   $$1=c_{1}e^{-1}+c_{2}e\,{\text{ and }}\,-1={\frac {1}{2}}(c_{1}e^{-1}-c_{2}e).$$
3. Solving these two equations gives: $c_{1}={\frac {-1}{2}}e,c_{2}={\frac {3}{2}}e^{-1}$ and so the solution for our IVP is:
   $$y\left(t\right)=-{\frac {1}{2}}e^{1+{\frac {t}{2}}}+{\frac {3}{2}}e^{-{\frac {t}{2}}-1}.$$
4. Therefore, as $t\to +\infty$ we obtain $y\to -\infty$.

• Consider the IVP

$$y''+5y'+6y=0,\,\,y\left(0\right)=2,\,\,y'\left(0\right)=\beta$$

1. The characteristic equation is $r^{2}+5r+6=0\Rightarrow r=-2,-3$ and so the general solution will be:
   $$y(t)=c_{1}e^{-2t}+c_{2}e^{-3t}$$
2. Using the initial conditions we obtain:
   $$2=c_{1}+c_{2}\,{\text{ and }}\,\beta =-2c_{1}-3c_{2}.$$
3. Solving these two equations gives: $c_{1}=(6+\beta ),\,c_{2}=-\left(4+\beta \right)$ and so the solution for our IVP is:
   $$y(t)=(6+\beta )e^{-2t}-(4+\beta )e^{-3t}.$$
4. Therefore, as $t\to +\infty$ we obtain $y\to 0$.

Second order equations

In order to show how these equations are solved, lets start with the most basic case - a second order equation

$A{\frac {d^{2}y}{dx^{2}}}+B{\frac {dy}{dx}}+Cy=0$

where A, B, and C are constants.

The Auxiliary Quadratic

For a DE

$a{\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=0,$

Make the following substitution:

$y=e^{mx}\,$

This gives also

${\frac {dy}{dx}}=me^{mx}\,$

${\frac {d^{2}y}{dx^{2}}}=m^{2}e^{mx}\,$

The DE is now

$am^{2}e^{mx}+bme^{mx}+ce^{mx}=0\,$

Dividing by $e^{mx}$ gives (note:$e^{mx}$ can never equal zero)

$am^{2}+bm+c=0\,$

This is the auxiliary quadratic (AQ) of the DE. There are four classes of outcomes to the auxiliary quadratic:

1. $b^{2}-4ac>0,\,$, giving two distinct, real, roots.
2. $b^{2}-4ac=0,\,$, giving two coincident real, roots.
3. $b^{2}-4ac<0,\,$, giving complex roots.

a: Purely imaginary roots.

b: Complex-conjugate pair.

The method of solution of the DE depends on the class of AQ.

Class 1: Distinct, Real Roots

Consider the DE

${\frac {d^{2}y}{dx^{2}}}+{\frac {dy}{dx}}-6y=0$

The AQ is

$m^{2}+m-6=0\,$

$(m+3)(m-2)=0\,$

This gives us the following roots:

$m=2,\ m=-3$

Going back to the substitution we made to obtain the AQ, we have

$y=e^{2x},\ y=e^{-3x}$

as two distinct solutions to the DE. According to the [[../Higher 1#Superposition_principle|superposition principle]], the general solution is therefore

$y=Ae^{2x}+Be^{-3x}\,$

General Solution to Class 1 DEs

Generalizing, for the Second Order DE

$a{\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=0$

with the auxiliary quadratic given by

$am^{2}+bm+c=0\,$

with roots α and β, the general solution is

$y=Ae^{\alpha x}+Be^{\beta x}\,$

Class 2: Coincident, Real Roots

Consider the DE

${\frac {d^{2}y}{dx^{2}}}-4{\frac {dy}{dx}}+4y=0$

The AQ is

$m^{2}-4m+4=0\,$

$(m-2)(m-2)=0\,$

so, $e^{2x}\,$ is a solution. However, we cannot have it as both solutions as the factor of two produced will be absorbed into the constant, leaving us with only one constant, and therefore a DE without a full solution.

For the other solution we will use the Method of Reduction of Order. To do so we assume that it is in the form of:

$y_{2}=u(x)y_{1}=u(x)e^{2x}\,$

At the end we will check if our assumption is correct. We will now substitute this equation and solve for u(x)

$y_{2}''-4y_{2}'+4y_{2}=0\,$

$u''(x)e^{2x}+2u'(x)e^{2x}+2u'(x)e^{2x}+4u(x)e^{2x}-4(u'(x)e^{2x}+2u(x)e^{2x})+4u(x)e^{2x}=0\,$

$u''(x)e^{2x}=0\,$

$e^{2x}\,$ is always non-zero so the only way the product can equal zero is if:

$u''(x)=0\,$

Integrating twice offers

$u(x)=a_{1}x+a_{2}\,$

Therefore

$y_{2}=(a_{1}x+a_{2})y_{1}\,$

$y_{2}=(a_{1}x+a_{2})e^{2x}\,$

Our general solution is:

$y=C_{1}y_{1}+C_{2}y_{2}\,$

$y=C_{1}e^{2x}+C_{2}(a_{1}x+a_{2})e^{2x}\,$

$y=(C_{1}+a_{2})e^{2x}+C_{2}a_{1}xe^{2x}\,$

Because each constant is arbitrary we can simply write

$y=C_{1}e^{2x}+C_{2}xe^{2x}\,$

The Method of Reduction of Order can be used on different equations and u(x) does not always equal x. You can see below that $y=xe^{x}$ is a valid solution.

$y=xe^{2x}\,$

${\frac {dy}{dx}}=2xe^{2x}+e^{2x}=e^{2x}(2x+1)\,$

${\frac {d^{2}y}{dx^{2}}}=2e^{2x}+2e^{2x}(2x+1)=e^{2x}(4x+4)\,$

To check substitute these into the original DE:

$e^{2x}(4x+4)-4e^{2x}(2x+1)+4xe^{2x}=0\,$

$4xe^{2x}+4e^{2x}-8xe^{2x}-4e^{2x}+4xe^{2x}=0\,$

$0=0\,$

Therefore, $y=xe^{2x}\,$ is a solution as well.

General Solution to Class 2 DEs

Generalizing, for the Second Order DE

$a{\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=0$

with the coincident root α of the AQ, the general solution is

$y=(A+Bx)e^{\alpha x}\,$

Class 3a: Purely Imaginary Roots

To have complex roots, the AQ must have a discriminant less than zero, so

$b^{2}-4ac<0\,$

Also, for the solution to be purely imaginary, the value of b must be exactly zero.

Therefore,

$4ac>0\,$

This means that a and c have to have the same signs: either a and c are both positive or they are both negative. If we consider our general second-order DE:

$a{\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=0$

Setting b to zero gives

$a{\frac {d^{2}y}{dx^{2}}}+cy=0$

Dividing through by a gives

${\frac {d^{2}y}{dx^{2}}}+{\frac {c}{a}}y=0$.

Therefore, the y term is always positive, and this can be represented by

${\frac {d^{2}y}{dx^{2}}}+\omega ^{2}y=0$.

(I'm using ω here as it is used for simple harmonic motion, which is the primary use of this DE). There are now two paths to the solution of the DE. The first relies on us spotting that we can use the cyclical nature of trig. functions when derived. Substitute the following

$y=\cos \omega x\,$

${\frac {dy}{dx}}=-\omega \sin \omega x\,$

${\frac {d^{2}y}{dx^{2}}}=-\omega ^{2}\cos \omega x\,$

And check in our DE:

${\frac {d^{2}y}{dx^{2}}}+\omega ^{2}y=-\omega ^{2}\cos \omega x+\omega ^{2}\cos \omega x=0$

This checks out, so $\cos \omega x\,$ is a solution. A similar result holds true for the substitution using $y=\sin \omega x\,$.

Our solutions are therefore

$y=\cos \omega x\,$

$y=\sin \omega x\,$

So the general solution is

$y=A\cos \omega x+B\sin \omega x.\,$

The other method of solving this equation is to use Euler's Formula:

$e^{\omega ix}=\cos \omega x+i\sin x\,$

and

$e^{-\omega ix}=\cos \omega x-i\sin x\,$

From our original DE, we have an AQ of

$m^{2}+\omega ^{2}=0\,$

giving us roots of

$m=\pm i\omega \,$

so the general solution, similar to the Class 1 DEs, is

$y=Ae^{\omega ix}+Be^{-\omega ix}\,$

$y=A(\cos \omega x+i\sin \omega x)+B(\cos \omega x-i\sin \omega x)\,$

$y=(A+B)\cos \omega x+i(A-B)\sin \omega x\,$

Since A and B are arbitrary, we can set new constants for convenience, letting our new A equal A+B and our new B equal i(A-B).

Thus we have as our general solution

$y=A\cos \omega x+B\sin \omega x\,$

General Solution to Class 3a DEs

Generalizing, for the Second Order DE

${\frac {d^{2}y}{dx^{2}}}+\omega ^{2}y=0$

the general solution is

$y=A\cos \omega x+B\sin \omega x\,$

Class 3b: Complex Conjugate Roots

Since it is a proven theorem that complex roots of polynomials always occur in conjugate pairs, the only remaining class of AQ is the one with complex conjugates for solutions.

Given that the solutions are complex, we know that in the AQ

$am^{2}+bm+c=0\,$

$b^{2}-4ac,<0,\ b\neq 0\,$ (see Class 3a).

The roots of this are in the form

$(p\pm iq)\,$

The general solution is then

$y=Ae^{(p+iq)x}+Be^{(p-iq)x}\,$

$y=e^{px}\left(Ae^{iqx}+Be^{-iqx}\right)\,$

From Euler's Formulas, we can now get

$y=e^{px}\left(A\left(\cos qx+i\sin qx\right)+B\left(\cos qx-i\sin qx\right)\right)\,$

$y=e^{px}\left((A+B)\cos qx+i(A-B)\sin qx\right)\,$

As A and B are arbitrary, we can collapse them as in Class 3a, so that we have the general solution

$y=e^{px}\left(A\cos qx+B\sin qx\right)\,$

General Solution to Class 3b DEs

Generalizing, for the Second Order DE

$a{\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=0$

with an AQ with roots

$m=p\pm iq,$

The general solution is

$y=e^{px}\left(A\cos qx+B\sin qx\right)\,$

We have now covered all possible types of homogeneous second-order differential equation, and we didn't even have to integrate anything! We will now have a look at higher order equivalents.

Licensing

Content obtained and/or adapted from:

• Homogeneous differential equation, Wikipedia under a CC BY-SA license
• Homogeneous second order equations, Wikibooks: Ordinary Differential Equations under a CC BY-SA license

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