Discriminant equal to zero

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

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: $\frac{\dd^2y}{\dd t^2}-6\frac{\dd y}{\dd t}+9\, y(t)=0,\qquad y(0)=-1, \qquad \frac{dy}{dt}(0)=3$

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 $y(0)=-1, \qquad \frac{\dd y}{\dd t}(0)=3$ 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 $y'(0)=3$ gives the equation $B+3A=3.$ So we must solve the following system of equations in the unknowns $A$ and $B$ : $\left\{\begin{aligned}A &= -1\\ B+3A&= 3\end{aligned}\right.$ The solution of this system of equations is $A=-1 \qquad\text{and}\qquad B=6$ So the solution of the initial value problem is $y(t)=\left(6\,t-1\right)\e^{3\,t}$

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