### Differential equations: Linear second-order differential equations

### Solving homogeneous linear ODEs with constant coefficients

Homogeneous linear differential equations of second order with constant coefficients can be solved fully.

General solution of a homogeneous linear second-order ODE with constant coefficients

Consider the ODE \[a\cdot\frac{\dd^2y}{\dd t^2}+b\cdot\frac{\dd y}{\dd t}+c\cdot y(t)=0\] where #a#, #b#, and #c# are constants with #a\ne0#. The corresponding *characteristic equation* is \[a\cdot\lambda^2+b\cdot\lambda+c=0\] The *discriminant* of this quadratic equation in #\lambda# is \(D=b^2-4a\cdot c\). We distinguish three cases according to the sign of #D# and give the general solution with integration constants #A# and #B# in each case:

- If #D# is positive, then the characteristic equation of the
*ABC*-formula two real solutions, say \[\lambda_1=\frac{-b+\sqrt{D}}{2a}\qquad\text{and}\qquad \lambda_2=\frac{-b-\sqrt{D}}{2a}\] The general solution of the differential equation is then \[ y(t)=A\cdot\e^{\lambda_1\cdot t}+B\cdot\e^{\lambda_2\cdot t}\] - If \(D\) is equal to zero, then, in accordance with the
*abc*-formula, the characteristic equation has exactly one solution, namely \[\lambda=\frac{-b}{2a}\] The general solution is then \[y(t)=(A+B\cdot t)\cdot\e^{\lambda\cdot t}\] - If \(D\) is negative, then the characteristic equation has no real solutions, but two complex conjugate solutions #\alpha\pm\beta\, \ii#, where \[\alpha=-\frac{b}{2a}\text{ and } \beta=\frac{\sqrt{-D}}{2a}\] The general solution of the differential equation is then \[y(t)=\e^{\alpha\cdot t}\cdot\bigl(A\cdot\cos(\beta \cdot t)+B\cdot\sin(\beta \cdot t)\bigr)\]

Solve the following differential equation: \[\frac{\dd^2y}{\dd t^2}+\frac{\dd y}{\dd t}-2\, y(t)=0\]

Use #c# and #d# as constants.

Use #c# and #d# as constants.

Solution #y(t)=# #c\cdot\e^{-2t}+d\cdot\e^{t}#

The

The discriminant \(D\) of the quadratic polynomial is equal to \[D = 1 ^ 2-4 \cdot (-2) = 9 = 3 ^ 2 \] Since #D\gt0#, the characteristic equation has two real solutions:\[\lambda_{1,2}= \frac{-1\pm 3}{2}\] that is, \[\lambda_1=-2 \qquad\text{and}\qquad\lambda_2=1\]

According to theorem

The

*characteristic equation*of the given homogeneous linear differential equation of second order is \[ \lambda^2+\lambda-2=0\]The discriminant \(D\) of the quadratic polynomial is equal to \[D = 1 ^ 2-4 \cdot (-2) = 9 = 3 ^ 2 \] Since #D\gt0#, the characteristic equation has two real solutions:\[\lambda_{1,2}= \frac{-1\pm 3}{2}\] that is, \[\lambda_1=-2 \qquad\text{and}\qquad\lambda_2=1\]

According to theorem

*General solution of the homogeneous linear second-order ODE with constant coefficients*, the differential equation has the following general solution: \[y(t)=c\cdot\e^{-2t}+d\cdot\e^{t}\] with constants \(c\) and \(c\).
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