Question

Find the real-valued solution to the initial value problem {y′1y′2==−3y1−2y2,5y1+3y2,y1(0)=3,y2(0)=−10. {y1′=−3y1−2y2,y1(0)=3,y2′=5y1+3y2,y2(0)=−10. use tt as the independent variable in your answers.

Answer 1

Given that

$y'_1=-3y_1-2y_2 \\ \\ y'_2=5y_1+3y_2 \ \ \ \ \ y_1(0)=3,\ \ y_2(0)=-10$

Then:

$\left|\begin{array}{cc}-3-\lambda&-2\\5&3-\lambda\end{array}\right| =0 \\ \\ (-3-\lambda)(3-\lambda)+10=0 \\ \\ \lambda^2+1=0 \\ \\ \lambda=\pm i$

The eigenvectors are as follows:

$\left(\begin{array}{cc}-3-i&-2\\5&3-i\end{array}\right)\left(\begin{array}{c}v_1\\v_2\end{array}\right) =\left(\begin{array}{c}0\\0\end{array}\right) \\ \\ \Rightarrow(-3-i)v_1-2v_2=0 \\ \\ \Rightarrow v_2= \frac{1}{2} (-3-i)v_1 \\ \\ \bold{\overrightarrow{v_1}=\left(\begin{array}{c}v_1\\\frac{1}{2} (-3-i)v_1\end{array}\right) =\left(\begin{array}{c}2\\-3-i\end{array}\right)}$

Therefore, the solution is given as follows:

$e^{\lambda t}v=e^{it}\left(\begin{array}{c}2\\-3-i\end{array}\right) \\ \\ =\left(\cos(t)+i\sin(t)\right)\left(\begin{array}{c}2\\-3-i\end{array}\right) \\ \\ =\left(\begin{array}{c}2\cos(t)+2i\sin(t)\\-3\cos(t)-3i\sin(t)-i\cos(t)+\sin(t)\end{array}\right) \\ \\ =\left(\begin{array}{c}2\cos(t)\\-3\cos(t)+\sin(t)\end{array}\right)+i\left(\begin{array}{c}2\sin(t)\\-3\sin(t)-\cos(t)\end{array}\right)$

Thus, the solution is given by:

$\left(\begin{array}{c}y_1\\y_2\end{array}\right)=c_1\left(\begin{array}{c}2\cos(t)\\-3\cos(t)+\sin(t)\end{array}\right)+c_2\left(\begin{array}{c}2\sin(t)\\-3\sin(t)-\cos(t)\end{array}\right) \\ \\ =\left(\begin{array}{c}2c_1\cos(t)+2c_2\sin(t)\\-3c_1\cos(t)+c_1\sin(t)-3c_2\sin(t)-c_2\cos(t)\end{array}\right)$

Applying the initial condition, we have

$2c_1\cos(0)+2c_2\sin(0)=3 \\ \\ \Rightarrow c_1= \frac{3}{2} \\ \\ -3c_1\cos(0)+c_1\sin(0)-3c_2\sin(0)-c_2\cos(0)=-10 \\ \\ \Rightarrow-3c_1-c_2=-10 \\ \\ \Rightarrow c_2=- \frac{9}{2} +10= \frac{11}{2}$

Therefore, the solution to the initial value problem is:

$y_1=3\cos(t)+11\sin(t) \\ \\ y_2=- \frac{9}{2} \cos(t)+\frac{3}{2}\sin(t)-\frac{33}{2}\sin(t)-\frac{11}{2}\cos(t) \\ \\ =-10\cos(t)-15\sin(t)$