Homogeneous Linear Differential Equations

A homogeneous linear differential equation is a differential equation in which every term is of the form \(y^{(n)}p(x)\) i.e. a derivative of \(y\) times a function of \(x\). In general, these are very difficult to work with, but in the case where all the constants are coefficients, they can be solved exactly. In this case, the differential equation looks like \[a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,\] with \(a_{0}, a_{1} ,a_{n-1} ,\ldots , a_{n}\) being real constants, and almost resembles a polynomial. In fact, looking at the roots of this associated polynomial gives solutions to the differential equation.

A reasonable guess for a solution to \(a_ny^{(n)}+\cdots+a_0y\) is \(y=e^{rx}\) - this is called an ansatz. Note that \[ \dfrac{d^ky}{dx^k} e^{rx} = r^k e^{rx},\] so for \(e^{rx}\) to be a solution, it would have to satisfy \[ a_{n} r^n e^{rx} + a_{n-1} r^{n-1} e^{rx} + \cdots + a_{1} r e^{rx} + a_{0} e^{rx} = 0.\] Since \( e ^ {rx} \neq 0 \), it must be the case that \[ a_ {n}r^n + a_ {n-1}r^{n-1}+ \cdots + a_ {1} r + a_ {0} = 0. \qquad (3).\] Thus, \(e^{rx}\) solves the differential equation when \(r\) is a root of this polynomial.

For the differential equation \[a_{n} \dfrac{d^ny}{dx^n} + a_{n-1} \dfrac{d^{n-1}y}{dx^{n-1}} + \cdots + a_{1} \dfrac{dy}{dx} + a_{0} y = 0,\] the associated characteristic equation is \(a_nr^n+\cdots+a_1r+a_0=0\).

Depending on the nature of the roots of the characteristic polynomial, the differential equation has slightly different solutions.

When the roots are real and distinct, the solutions are just the linear combinations of \(e^{rx}\) for the different roots \(r\).

If the \(n\) roots \(r_1, r_2, \ldots, r_{n}\) of the characteristic equation are real and distinct, then \[ y(x) = c_1 e^{r_1x} + c_2e^{r_2x} +\cdots +c_ne^{r_nx}\] is a general solution of the equation \((1),\) with \(c_1, c_2, \ldots, c_n\) constants.

Solve \[\]

\[\begin{array} &y'' + 2y' - 8y = 0, &y(0) = 5, &y'(0) = - 2.\end{array}\]

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We need to solve the characteristic equation \(r^2 + 2r - 8 = 0.\)

Observe that \(r = 2, -4\) are the roots of this equation. Then since \(y(x) = c_1 e^{2x} + c_2 e^{-4x}\) is the general solution, \(y'(x) = 2 c_1 e^{2x} - 4 c_2 e^{-4x}\). Now, using the initial conditions, we have \[y(0) = c_1 + c_2 = 5, y'(0) = 2c_1 - 4c_2 = -2 \Rightarrow c_1 = 3, c_2=2.\]

Thereofore, the desired particular solution is \[y(x) = 3e^{2x} + 2e^{-4x}. \ _\square\]

When the characteristic polynomial has repeated roots, the previous theorem no longer applies.

If the characteristic equation has a repeated real root \(r\) of multiplicity \(k,\) then part of the general solution of the differential equation corresponding to \(r\) in equation is of the form \[(c_1 + c_ {2}x + c_ {3}x^2 + \cdots + c_ {k}x^{k-1}) \cdot e^{rx}.\]

Find a general solution of \[\] \[y^{(4)} + 3 \cdot y^{(3)} + 3 \cdot y'' + y' = 0.\]

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The characteristic equation of the differential equation is \(r^4 + 3 \cdot r^3 + 3 \cdot r^2 + r= r \cdot (r+1)^3=0\).

It has the single root \(r_1 = 0,\) which gives the solution \(y_1 = c_1\) to the general solution, and the triple root \((k=3)\) \(r_2 = -1,\) which gives \(y_2 = (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}.\) Thus, the general solution of the differential equation is \[y(x) = c_1 + (c_2 + c_ {3}x + c_ {4}x^2) \cdot e^{-x}. \ _\square\]

Because the coefficients of the differential equation and its characteristic equation are real, any root complex appears in complex conjugate pair \(a \pm bi,\) where \(a\) and \(b\) are real and i = \(\sqrt{-1}.\)

If the characteristic equation has a pair of complex roots not repeated \(a \pm bi\), then the relevant part to them of the general solution of equation has the form \(e^{ax} \cdot (c_1 \cos bx + c_2 \sin bx)\).

Solve \(y'' - 4y' + 5y = 0.\)

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The characteristic equation is \(r^2 - 4r + 5 = 0,\) whose roots are \(2 \pm i.\) Thus, the general solution is \(y(x) = e^{2x} \cdot (c_1 \cos x + c_2 \sin x).\)

When there are repeated complex roots, they can be accounted for in the same way as with repeated real roots.

Find a general solution of \[y^{(4)} + 4 \cdot y^{(3)} + 12 \cdot y'' + 16y' + 16y = 0.\]

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The characteristic equation is \((r^2 + 2r + 4)^2 = 0,\) whose roots are \(-1 \pm i \cdot \sqrt{3}\) of multiplicity \(2.\) Therefore, the general solution is \[y(x) =e^{-x} \cdot \left(c_1 \cos x\sqrt{3} + d_1 \sin x\sqrt{3}\right) + xe^{-x} \cdot \left(c_2 \cos x\sqrt{3} + d_2 \sin x\sqrt{3}\right).\]