Question

Solve DE r^2 +4r+3=12

Answer 1

If I understood you correctly, then you have written next equation:

$$y^{||} + 4y^{|} + 3y = 12$$

Solution:

Let's write the characteristic equation of ODE:

$$x^{2} + 4x + 3 = 0$$ $$D = b^{2} - 4ac = 16 - 4 * 3 = 4$$ $$x_{1} = \frac{-b + \sqrt{D}}{2a} = \frac{-4 + 2}{2} = -1$$ $$x_{2} = \frac{-b - \sqrt{D}}{2a} = \frac{-4 - 2}{2} = -3$$

The general solution of the given differential equation is

$$y = y_{\text{particular}} + C_{1} * e^{-t} + C_{2} * e^{-3t}$$

Now find $y_{\text{particular}}$ which have such form:

$$y_{\text{particular}} = A$$ $$y'_{\text{particular}} = 0$$ $$y''_{\text{particular}} = 0$$

Now substitute these roots into the ODE:

$$0 + 4 * 0 + 3 * A = 12$$ $$A = 4$$

Now general solution is:

$$y = 4 + C_{1} * e^{-t} + C_{2} * e^{-3t}$$

Answer: $y = 4 + C_{1} * e^{-t} + C_{2} * e^{-3t}$