Question

The solution of the system of ODEs: \(\frac{dx}{dt}=x+2y\), \(\frac{dy}{dt}=3x+2y\) with initial conditions x(0)=6 and y(0)=4 is ____.

• A. \(x(t)=4e^{4t}+2e^t,\ y(t)=6e^{4t}-2e^t\)
• B. \(x(t)=4e^{4t}+2e^{-t},\ y(t)=6e^{4t}-2e^{-t}\)
• C. \(x(t)=6e^{4t}-2e^{-t},\ y(t)=2e^{4t}+2e^{-t}\)
• D. \(x(t)=2e^{4t}+4e^{-t},\ y(t)=2e^{4t}+2e^{-t}\)

Hint:

Find the eigenvalues of the coefficient matrix (4 and -1), form the general solution, then fit the constants to x(0)=6, y(0)=4.

Answer 1

The Correct Option is B

Step 1: From the first equation, \(\frac{dx}{dt}=x+2y\), solve for \(y\):

\[y = \frac{x'-x}{2}\]

Step 2: Differentiate with respect to \(t\):

\[y' = \frac{x''-x'}{2}\]

Step 3: Substitute both into the second equation \(y'=3x+2y\):

\[\frac{x''-x'}{2} = 3x + 2\cdot\frac{x'-x}{2} = 3x + x' - x = x' + 2x\]

Step 4: Multiply through by 2 and simplify:

\[x''-x' = 2x'+4x \implies x''-3x'-4x=0\]

Step 5: The characteristic equation is \(r^2-3r-4=0\), giving \((r-4)(r+1)=0\), so \(r=4,-1\). Hence

\[x(t) = C_1e^{4t}+C_2e^{-t}\]

Step 6: Recover \(y\) from Step 1: \(x'(t) = 4C_1e^{4t}-C_2e^{-t}\), so

\[y(t) = \frac{x'-x}{2} = \frac{(4C_1e^{4t}-C_2e^{-t})-(C_1e^{4t}+C_2e^{-t})}{2} = \frac{3C_1e^{4t}-2C_2e^{-t}}{2} = 1.5C_1e^{4t}-C_2e^{-t}\]

Step 7: Apply \(x(0)=6\): \(C_1+C_2=6\). Apply \(y(0)=4\): \(1.5C_1-C_2=4\). Adding gives \(2.5C_1=10\), so \(C_1=4\) and \(C_2=2\).

Step 8: Substitute back:

\[x(t) = 4e^{4t}+2e^{-t}, \qquad y(t) = 1.5(4)e^{4t}-2e^{-t} = 6e^{4t}-2e^{-t}\] \[\boxed{x(t)=4e^{4t}+2e^{-t},\ y(t)=6e^{4t}-2e^{-t}}\]