Go back to the home page

\begin{equation*} y” + 3y' = 0, \quad y(0) = -2, \quad y'(0) = 3 \end{equation*} Since this is a linear homogeneous constant-coefficient ODE, the solution is of the form \(y = e^{rt}\) \begin{equation*} y=e^{rt} \quad \implies \quad y' = re^{rt} \quad \implies \quad y” = r^2 e^{rt} \end{equation*} Substitute those expressions into the ODE \begin{equation*} r^2 e^{rt} + 3(re^{rt}) = 0 \end{equation*} Divide both sides by \(e^{rt}\) \begin{equation*} r^2 + 3r = 0 \end{equation*} Roots of this polynomial are \(r_0 = -3\) and \(r_1 = 0\). Two solutions to the ODE are \(y=e^{-3t}\) and \(y=e^0=1\). Therefore, the general solution is \begin{equation*} y(t) = C_1 e^{-3t} + C_2 \end{equation*} Differentiating \(y\) gives us \begin{equation*} y'(t) = -3C_1 e^{-3t} \end{equation*} Now, we can determine our constants by applying the two initial conditions we know \begin{equation*} \begin{cases} y(0) = C_1 + C_2 = -2\\ y'(0) = -3C_1 = 3 \end{cases} \end{equation*} Therefore \(C_1 = -1\) and \(C_2 = -1\), therefore \begin{equation*} y(t) = -e^{-3t} - 1 \end{equation*} This solution converges to \(-1\) as \(t \to \infty\).

Find a differential equation whose general solution is \begin{equation*} y=c_{1} e^{2 t}+c_{2} e^{-3 t} \end{equation*} We see the roots are \(r_0 = -3\) and \(r_1 = 2\). Alternatively, you can make a set of solutions, and call it \(r = \{-3,2\}\). So \begin{align*} (r+3)(r-2)&=0 \\ \implies r^2 + r - 6 &= 0 \end{align*} Multiply both sides by \(e^{rt}\) \begin{align*} r^2e^{rt} + re^{rt} - 6e^{rt} = 0 \end{align*} Therefore, the differential equation is \begin{align*} y” + y' - 6y = 0 \end{align*}

This is a linear homogeneous constant-coefficient ODE, apply the same method as in Problem 9. Find that \(r = \{-1, 2\}\) and the general solution is \begin{align*} y(t) = C_1 e^{-t} + C_2 e^{2t} \end{align*} The derivative would be \begin{align*} y'(t) = -C_1 e^{-t} + 2 C_2 e^{2t} \end{align*} Let us solve the initial conditions \begin{align*} \begin{cases} y(0) = C_1 + C_2 = \alpha\\ y'(0) = -C_1 + 2C_2 = 2 \end{cases} \implies \begin{cases} C_1 = \frac{2}{3}(\alpha -1)\\ C_2 = \frac{1}{3}(\alpha +2) \end{cases} \end{align*} Therefore, \begin{align*} y(t) = \frac{2}{3}(\alpha -1)e^{-t} + \frac{1}{3}(\alpha +2)e^{2t} \end{align*} We can see that if \(t \to \infty\), then \(y \to \infty\). Therefore, set \(\alpha = -2\).

\begin{align*} y” + 5y' + 6y = 9, \quad y(0) = 2, \quad y'(0) = \beta, \end{align*} where \(\beta > 0\).

This is a linear homogeneous constant-coefficient ODE, find that \(r = {-\frac{1}{2}, \frac{1}{2}}\). The two solutions are \begin{align*} y(t) = C_1 e^{-\frac{t}{2}} + C_2 e^{\frac{t}{2}} \end{align*} Then \begin{align*} y'(t) = -\frac{C_1}{2} e^{-\frac{t}{2}} + \frac{C_2}{2} e^{\frac{t}{2}} \end{align*} Solve \begin{align*} \begin{cases} y(0) = C_1 + C_2 = 2\\ y'(0) = -\frac{C_1}{2} + \frac{C_2}{2} = \beta \end{cases} \implies \begin{cases} C_1 = 1 - \beta\\ C_2 = 1 + \beta \end{cases} \end{align*} Finally, \begin{align*} y(t) = (1 - \beta)e^{-\frac{t}{2}} + (1+\beta)e^{\frac{t}{2}} \end{align*} To prevent the solution from going to the infinity and beyond, set \(\beta=-1\).

See Professor Van Vleck’s notes on this problem.

\begin{align*} ay” + by' + cy = 0, \end{align*} where \(a, b, c \in \mathbb{R}\) and \(a > 0\).

This is yet again another linear homogeneous constant-coefficient ODE. Find that \begin{align*} a\left(r^{2} e^{r t}\right)+b\left(r e^{r t}\right)+c\left(e^{r t}\right)=0 \end{align*} Divide both sides by \(e^{r t}\) \begin{align*} a r^{2}+b r+c=0 \\ \implies r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \end{align*}

For the roots to be real, different and negative, \(b>0\) and \(0 < c < \frac{b^2}{4a}\).

For the roots to be real with opposite signs, \(c < 0\).

For the roots to be real, different and positive, \(b<0\) and \(0 < c < \frac{b^2}{4a}\).

The Wronskian of these two functions is \begin{align*} W &=\left|\begin{array}{cc} \cos ^{2} \theta & 1+\cos 2 \theta \\ \frac{d}{d \theta}\left(\cos ^{2} \theta\right) & \frac{d}{d \theta}(1+\cos 2 \theta) \end{array}\right| \\ &=\left|\begin{array}{cc} \cos ^{2} \theta & 1+\cos 2 \theta \\ 2 \cos \theta(-\sin \theta) & -2 \sin 2 \theta \end{array}\right| \\ &=\cos ^{2} \theta(-2 \sin 2 \theta)-(1+\cos 2 \theta)[2 \cos \theta(-\sin \theta)] \\ &=-2 \cos ^{2} \theta \sin 2 \theta+2 \sin \theta \cos \theta(1+\cos 2 \theta) \\ &=-2 \cos ^{2} \theta(2 \sin \theta \cos \theta)+2 \sin \theta \cos \theta\left(1+2 \cos ^{2} \theta-1\right) \\ &=-4 \cos ^{2} \theta \sin \theta \cos \theta+4 \sin \theta \cos \theta \cos ^{2} \theta \\ &=0 \end{align*}

\begin{align*} y” - y' - 2y = 0 \end{align*}

Note: Solutions for this problem are based on Jock’s solutions.

Calculate \(W\left(y_{1}, y_{2}\right)\) the Wronskian of \(y_{1}\) and \(y_{2}\).

\begin{align*} W\left(y_{1}, y_{2}\right) &=\left|\begin{array}{ll} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{array}\right| \\ &=\left|\begin{array}{cc} e^{-t} & e^{2 t} \\ -e^{-t} & 2 e^{2 t} \end{array}\right| \\ &=e^{-t}\left(2 e^{2 t}\right)-e^{2 t}\left(-e^{-t}\right) \\ &=2 e^{t}+e^{t} \\ &=3 e^{t} \end{align*}

Since \(W\left(y_{1}, y_{2}\right) \neq 0, y_{1}\) and \(y_{2}\) form a fundamental set of solutions.

Check that \(y_{3}\) is a solution of the ODE.

\begin{align*} \begin{array}{c} y_{3}^{\prime \prime}-y_{3}^{\prime}-2 y_{3} \stackrel{?}{=} 0 \\ \frac{d^{2}}{d t^{2}}\left(-2 e^{2 t}\right)-\frac{d}{d t}\left(-2 e^{2 t}\right)-2\left(-2 e^{2 t}\right) \stackrel{?}{=} 0 \\ \left(-8 e^{2 t}\right)-\left(-4 e^{2 t}\right)-2\left(-2 e^{2 t}\right) \stackrel{?}{=} 0 \\ -8 e^{2 t}+4 e^{2 t}+4 e^{2 t} \stackrel{?}{=} 0 \\ 0=0 \end{array} \end{align*}

Now check that \(y_{4}=e^{-t}+2 e^{2 t}\) is a solution of the ODE.

\begin{align*} \begin{array}{c} y_{4}^{\prime \prime}-y_{4}^{\prime}-2 y_{4} \stackrel{?}{=} 0 \\ \frac{d^{2}}{d t^{2}}\left(e^{-t}+2 e^{2 t}\right)-\frac{d}{d t}\left(e^{-t}+2 e^{2 t}\right)-2\left(e^{-t}+2 e^{2 t}\right) \stackrel{?}{=} 0 \\ \left(e^{-t}+8 e^{2 t}\right)-\left(-e^{-t}+4 e^{2 t}\right)-2\left(e^{-t}+2 e^{2 t}\right) \stackrel{?}{=} 0 \\ e^{-^{\ell}}+8 e^{2 t}+e^{-}-4 e^{2 t}-2 e^{-}-4 e^{2 t} \stackrel{?}{=} 0 \\ 0=0 \end{array} \end{align*}

Now check that \(y_{5}=2 y_{1}(t)-2 y_{3}(t)=2 e^{-t}-2\left(-2 e^{2 t}\right)=2 e^{-t}+4 e^{2 t}\) is a solution of the ODE.

\begin{align*} \begin{array}{c} y_{5}^{\prime \prime}-y_{5}^{\prime}-2 y_{5} \stackrel{?}{=} 0 \\ \frac{d^{2}}{d t^{2}}\left(2 e^{-t}+4 e^{2 t}\right)-\frac{d}{d t}\left(2 e^{-t}+4 e^{2 t}\right)-2\left(2 e^{-t}+4 e^{2 t}\right) \stackrel{?}{=} 0 \\ \left(2 e^{-t}+16 e^{2 t}\right)-\left(-2 e^{-t}+8 e^{2 t}\right)-2\left(2 e^{-t}+4 e^{2 t}\right) \stackrel{?}{=} 0 \\ 2 e^{-}+16 e^{2 t}+2 e^{-}-8 e^{2 t}-4 e^{-}-8 e^{2 t} \stackrel{?}{=} 0 \\ 0=0 \end{array} \end{align*}

Calculate \(W\left(y_{1}, y_{3}\right)\), the Wronskian of \(y_{1}\) and \(y_{3}\).

\begin{aligned} W\left(y_{1}, y_{3}\right) &=\left|\begin{array}{ll} y_{1} & y_{3} \\ y_{1}^{\prime} & y_{3}^{\prime} \end{array}\right| \\ &=\left|\begin{array}{cc} e^{-t} & -2 e^{2 t} \\ -e^{-t} & -4 e^{2 t} \end{array}\right| \\ &=e^{-t}\left(-4 e^{2 t}\right)-\left(-2 e^{2 t}\right)\left(-e^{-t}\right) \\ &=-4 e^{t}-2 e^{t} \\ &=-6 e^{t} \end{aligned}

Since \(W\left(y_{1}, y_{3}\right) \neq 0, y_{1}\) and \(y_{3}\) form a fundamental set of solutions.

Now calculate \(W\left(y_{2}, y_{3}\right)\), the Wronskian of \(y_{2}\) and \(y_{3}\)

\begin{aligned} W\left(y_{2}, y_{3}\right) &=\left|\begin{array}{ll} y_{2} & y_{3} \\ y_{2}^{\prime} & y_{3}^{\prime} \end{array}\right| \\ &=\left|\begin{array}{cc} e^{2 t} & -2 e^{2 t} \\ 2 e^{2 t} & -4 e^{2 t} \end{array}\right| \\ &=e^{2 t}\left(-4 e^{2 t}\right)-\left(-2 e^{2 t}\right)\left(2 e^{2 t}\right) \\ &=-4 e^{4 t}+4 e^{4 t} \\ &=0 \end{aligned}

Since \(W\left(y_{2}, y_{3}\right)=0, y_{2}\) and \(y_{3}\) do not form a fundamental set of solutions. Now calculate \(W\left(y_{1}, y_{4}\right)\), the Wronskian of \(y_{1}\) and \(y_{4}\)

\begin{aligned} W\left(y_{1}, y_{4}\right) &=\left|\begin{array}{ll} y_{1} & y_{4} \\ y_{1}^{\prime} & y_{4}^{\prime} \end{array}\right| \\ &=\left|\begin{array}{cc} e^{-t} & e^{-t}+2 e^{2 t} \\ -e^{-t} & -e^{-t}+4 e^{2 t} \end{array}\right| \\ &=e^{-t}\left(-e^{-t}+4 e^{2 t}\right)-\left(e^{-t}+2 e^{2 t}\right)\left(-e^{-t}\right) \\ &=-e^{-2 t}+4 e^{t}+e^{-2 t}+2 e^{t} \\ &=6 e^{t} \end{aligned}

Since \(W\left(y_{1}, y_{4}\right) \neq 0, y_{1}\) and \(y_{4}\) form a fundamental set of solutions. Now calculate \(W\left(y_{4}, y_{5}\right)\), the Wronskian of \(y_{4}\) and \(y_{5}\).

\begin{aligned} W\left(y_{4}, y_{5}\right) &=\left|\begin{array}{ll} y_{4} & y_{5} \\ y_{4}^{\prime} & y_{5}^{\prime} \end{array}\right| \\ &=\left|\begin{array}{cc} e^{-t}+2 e^{2 t} & 2 e^{-t}+4 e^{2 t} \\ -e^{-t}+4 e^{2 t} & -2 e^{-t}+8 e^{2 t} \end{array}\right| \\ &=\left(e^{-t}+2 e^{2 t}\right)\left(-2 e^{-t}+8 e^{2 t}\right)-\left(2 e^{-t}+4 e^{2 t}\right)\left(-e^{-t}+4 e^{2 t}\right) \\ &=-2 e^{-2 t}+8 e^{t}-4 e^{t}+16 e^{4 t}-\left(-2 e^{-2 t}+8 e^{t}-4 e^{t}+16 e^{4 t}\right) \\ &=0 \end{aligned}

Since \(W\left(y_{4}, y_{5}\right)=0, y_{4}\) and \(y_{5}\) do not form a fundamental set of solutions.

\begin{align*} (\cos t)y” +(\sin t)y' -ty = 0 \end{align*} Then \begin{align*} y” + \frac{\sin t}{\cos t} - \frac{t}{\cos t}y = 0 \end{align*} so \begin{align*} p(t) = \tan t \end{align*} Then \begin{align*} W = C \exp\left(-\int \tan t dt \right) \end{align*} By Abel’s Theorem \begin{align*} W = C \exp \left( \ln (cos t) \right) \implies W = C\times \cos t \end{align*}

The equation \begin{align*} P(x) y^{\prime \prime}+Q(x) y^{\prime}+R(x) y=0 \end{align*} is said to be exact if it can be written in the form \begin{align*} \left(P(x) y^{\prime}\right)^{\prime}+(f(x) y)^{\prime}=0 \end{align*} where \(f(x)\) is to be determined in terms of \(P(x), Q(x)\), and \(R(x)\) The latter equation can be integrated once immediately, resulting in a first-order linear equation for \(y\) that can be solved as in Section 2.1. By equating the coefficients of the preceding equations and then eliminating \(f(x)\), show that a necessary condition for exactness is \begin{align*} P^{\prime \prime}(x)-Q^{\prime}(x)+R(x)=0 \end{align*} It can be shown that this is also a sufficient condition.