Q3RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is;

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{1}{2}\mathbf{t}}\mathbf{cos}\left(\frac{3t}{2}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{1}{2}\mathbf{t}}\mathbf{sin}\left(\frac{3t}{2}\right)$

Step 1: The general solution if the roots are complex conjugate

If the roots of the auxiliary equation are$\alpha \pm i\beta$, then the general solution is given as $y\left(t\right)=e^{\alpha t}\left(c_{1}cos\beta t+c_{2}sin\beta t\right).$

Step 2: Write the auxiliary equation of the given differential equation

The given differential equation is$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation$\mathbf{4}{\mathbf{m}}^2\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{10}\mathbf{=}\mathbf{0}.$

 

Step 3: Find the roots of the auxiliary equation

 Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{4}{\mathbf{m}}^2\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{10}\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\left(-4\right)\mathbf{\pm }\sqrt{16-160}}{\mathbf{2}\left(4\right)}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{4}\mathbf{\pm }\sqrt{-144}}{8}\\ \mathbf{m}\mathbf{=}\frac{4\pm 12i}{8}\\ \mathbf{m}\mathbf{=}\frac{1\pm 3i}{2}\end{array}$

The roots of the auxiliary equation are${\mathbf{m}}_1\mathbf{=}\frac{1+3i}{2}\&{\mathbf{m}}_2\mathbf{=}\frac{1-3i}{2}.$

Thus, the general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{1}{2}\mathbf{t}}\mathbf{cos}\left(\frac{3t}{2}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{1}{2}\mathbf{t}}\mathbf{sin}\left(\frac{3t}{2}\right).$