Discriminant equal to zero

Ordinary differential equations: Second-order linear ODEs with constant coefficients

Discriminant equal to zero

Characteristic equation with discriminant equal to zero We consider the homogeneous second-order linear differential equation with constant coefficients, written in the form \[a\,\frac{\dd^2y}{\dd t^2}+b\,\frac{\dd y}{\dd t}+c\,y(t)=0\] The corresponding characteristic equation is \[a\,\lambda^2+b\,\lambda+c=0\]

We restrict ourselves to the case where the discriminant \(D=b^2-4ac\) is equal to zero. Then there is according to the abc-formula only one real root, namely \[\lambda=\frac{-b}{2a}\] The following function is a solution of differential equation \[y_1(t)=e^{\lambda t}\] But are there perhaps more solution? Yes! \[y_2(t)=t e^{\lambda t}\] The general solution of the differential equation can be written as \[\begin{aligned}y(t)&=c_1\,y_1(t)+c_2\,y_2(t)\\\\ &=c_1\,e^{\lambda t}+c_2\, t e^{\lambda t}\\\\ &=(c_1+c_2 t) e^{\lambda t}\end{aligned}\] with constants \(c_1\) and \(c_2\). These constants are determined by boundary conditions.

The proof that \(y_2(t)=t\, e^{\lambda t}\) is also a solution of the differential equation is not difficult: \[\frac{\dd y_2}{\dd t}=(1+\lambda t)e^{\lambda t}\qquad\text{and}\qquad \frac{\dd^2y_2}{\dd t^2}=(\lambda t+2)\lambda e^{\lambda t}\] So \[\begin{aligned}a\,\frac{\dd^2y_2}{\dd t^2}+b\,\frac{\dd y_2}{\dd t}+c\,y_2(t) &= a (\lambda t+2)\lambda e^{\lambda t}+b(1+\lambda t)e^{\lambda t}+c t e^{\lambda t}\\ \\ &= \bigl((a\lambda^2+b\lambda+c)t+(2a\lambda+b)\bigr) e^{\lambda t}\\ \\ &= -\frac{b^2-4ac}{4a} t e^{\lambda t}\qquad\text{because }\lambda=\frac{-b}{2a}\\ \\ &= 0\qquad\text{because }b^2-4ac=0\end{aligned}\]

Solve the following initial value problem: \[\frac{\dd^2y}{\dd t^2}+6\frac{\dd y}{\dd t}+9\, y(t)=0,\qquad y(0)=1, \qquad \frac{dy}{dt}(0)=-2\]

The discriminant \(D\) of characteristic polynomial \[\lambda^2+6\lambda+9=0\] is equal to \[D=(6\times 6)-(4\times 9) = 0 \] There is one root of the characteristic polynomial, namely \[\lambda= \frac{-6}{2}=-3\] If you had found the equality \[\lambda^2+6\lambda+9=(\lambda+3)^2\] through factorization of a quadratic equation by inspection or a special product had recognized, then this conclusion would have been obvious .

Thus, the differential equation has the following general solution: \[y(t)=(A+B\,t)\,e^{-3t}\] with constants \(A\) and \(B\).

Using the initial values \[y(0)=1, \qquad \frac{\dd y}{\dd t}(0)=-2\] we can define equations that \(A\) and \(B\) must satisfy. Substitution of \(y(0)=1\) in the general solution \(y(t)=(A+B\,t)\,e^{-3t}\) gives \[A=1\] For the usage of the second initial value we need the derivative of the general solution; this can be computed with the calculation rules for differentiating and the derivative of the exponential function: \[y'(t)=B\,e^{-3t}-3(A+B\,t)\,e^{-3t}\] Substitution of \(y'(0)=-2\) gives the equation \[B-3A=-2.\] So we must solve the following system of equations in the unknowns \(A\) and \(B\): \[\left\{\begin{aligned}A &= 1\\ B-3A&= -2\end{aligned}\right.\] The solution of this system of equations is \[A=1 \qquad\text{and}\qquad B=1\] So the solution of the initial value problem is \[y(t)=\left(t+1\right)\e^ {- 3\,t }\]

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