Differential Equations

Have a go first. Then open the working one line at a time.

1. Matrix form

   \(\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}2&1\\3&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\)

   Row 1 holds the coefficients of x and y in dx/dt. Row 2 holds those in dy/dt. x' means dx/dt.

2. Characteristic equation

   \(\det(\mathbf M-\lambda\mathbf I)=(2-\lambda)(4-\lambda)-(1)(3)=0\)

   Eigenvalues are the values of λ for which Mv = λv has a non-zero solution v.

3. Tidy

   \(\lambda^2-6\lambda+5=0\)

   The 6 is the trace (2 + 4) and the 5 is the determinant (8 − 3).

4. Solve

   \((\lambda-1)(\lambda-5)=0\Rightarrow\lambda=1\ \text{or}\ 5\)

   From the calculator: eigenvalues 1 and 5. Use it to check.

Have a go first. Then open the working one line at a time.

1. Eigenvector for λ = 1

   \((\mathbf M-1\mathbf I)\mathbf v=\mathbf 0:\quad x+y=0\ \Rightarrow\ \mathbf p_1=\begin{pmatrix}1\\-1\end{pmatrix}\)

   One equation is enough: the two rows say the same thing. Any non-zero multiple of the vector works.

2. Eigenvector for λ = 5

   \((\mathbf M-5\mathbf I)\mathbf v=\mathbf 0:\quad -3x+y=0\ \Rightarrow\ \mathbf p_2=\begin{pmatrix}1\\3\end{pmatrix}\)

   One equation is enough: the two rows say the same thing. Any non-zero multiple of the vector works.

3. Check

   \(\mathbf M\mathbf p_2=\begin{pmatrix}2+3\\3+12\end{pmatrix}=5\begin{pmatrix}1\\3\end{pmatrix}\)

   Use the second row to check the eigenvector.

4. General solution

   \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{5t}\begin{pmatrix}1\\3\end{pmatrix}\)

   A and B are two arbitrary constants. Each eigenvalue goes in its own exponent.

Have a go first. Then open the working one line at a time.

1. Eigenvectors

   \(\lambda=-2:\ \mathbf p_1=\begin{pmatrix}1\\-1\end{pmatrix};\quad \lambda=3:\ \mathbf p_2=\begin{pmatrix}1\\4\end{pmatrix}\)

   For λ = 3: (−1 − 3)x + y = 0 gives y = 4x.

2. General solution

   \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-2t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{3t}\begin{pmatrix}1\\4\end{pmatrix}\)

   Write out both rows to get x and y separately.

3. Put t = 0

   \(A+B=7,\quad -A+4B=3\)

   e⁰ = 1, so the exponentials disappear. One equation for x(0), one for y(0).

4. Solve for A and B

   \(A=5,\ B=2\)

   Solve for A and B: add the equations to get 5B = 10.

5. Particular solution

   \(x=5e^{-2t}+2e^{3t},\quad y=-5e^{-2t}+8e^{3t}\)

   Check: x(0) = 5 + 2 = 7 and y(0) = −5 + 8 = 3.

Have a go first. Then open the working one line at a time.

1. Find y(0)

   \(x'(0)=3x(0)+2y(0):\ 17=15+2y(0)\Rightarrow y(0)=1\)

   Put t = 0 in dx/dt = 3x + 2y. This turns the x' condition into a y(0) condition.

2. Eigenvectors

   \(\lambda=1:\ \mathbf p_1=\begin{pmatrix}1\\-1\end{pmatrix};\quad\lambda=4:\ \mathbf p_2=\begin{pmatrix}2\\1\end{pmatrix}\)

   λ = 4: (3 − 4)x + 2y = 0 gives x = 2y.

3. General solution

   \(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}1\\-1\end{pmatrix}+Be^{4t}\begin{pmatrix}2\\1\end{pmatrix}\)

   x = A e^t + 2B e^{4t} and y = −A e^t + B e^{4t}.

4. Put t = 0

   \(A+2B=5,\quad -A+B=1\ \Rightarrow\ A=1,\ B=2\)

   Solve for A and B (from the calculator, or add the equations).

5. Answer

   \(x(t)=e^{t}+4e^{4t}\)

   Check: x'(0) = 1 + 16 = 17.