Which of the following equations are linear? For the ones that aren’t, explain where the problem lies. For the ones that are, put them in linear normal form if they aren’t already.
\(y' = ty- 7\)
\(ty' + y^2 = e^t\)
\(yy' = 5t - \sin(y)\)
\(\frac{y'}{t} +\tan(t)y=1\)
\(y' + \frac{t}{y} = t^2\)
\[\sin(t)y' - \frac{\log(t^2) + \frac{y}{t}}{1+e^{2t}}=y'\]
For #2–#5, find a general solution for the following homogeneous differential equations.
\(\dot{y} + y = 0\)
\(tw' + 2w = 0\)
\(\cos(x) y' = -\sin(x) y\)
\((t+1)y' + 3y = 0\)
For #6 – #11, find a general solution for the following nonhomogeneous differential equations.
\(w' - 4w = 7\)
\(\dot{y} + \frac{1}{t} \cdot y = 3\)
\(y' - 2y = xe^{2x}\)
\(y' + y = (1+t)\sin(t) - ty'\)
\(t\log(t) z' + z = 2\log(t)\)
\(y' + p(t)y = 1\)
For #12 – #18, Solve the following initial value problems. What is the largest interval containing the initial conditions on which your solution is defined?
\(y' + 3y = e^t\), \(y(0) = 4\)
\[\frac{\dee z}{\dt} + \frac{z}{\sin(t)} = \frac{1}{t^2-25} \text{ where } z(4) = 2\]
\[\tan(t) y' - y = \frac{1}{1-t} \text{ where } y(2) = 1\]
\(y' + 2ty = e^{-t^2}\), \(y(0) = 1\)
\(\displaystyle \theta' + \frac{1}{t}\,\theta = \frac{1}{t^2 - 4}\), \(\theta(\sqrt{5}) = 2\)
\[x' - x/(e^{-t}-1) = \tan(t) \text{ where } x(1) = 3\]
\(\displaystyle \frac{y'}{\cos(t)} + y = 1\), \(y(\pi) = 3\)
\(\displaystyle v' + \frac{1}{t-1}\,v = \frac{1}{t^3 - t}\), \(v(\frac{1}{2}) = -4\)
This exercise is a check that the recipe given does indeed give solutions to first order linear differential equations. The equation we want to verify is \[y'(t) + a(t)y(t) = b(t)\text{,}\] and the solution the recipe gives us, for an arbitary constant \(c\), is \[y(t) = e^{-\int a(t)\,dt} \int b(t) e^{ \int a(t)\,dt}\,dt + ce^{-\int a(t)\,dt}\,\text{.}\] [Hint. If you’re a go-getter, it is possible to check this by computing \(y'\) and \(ay\), then adding them to see that everything cancels except \(b\). It might be less onerous if you define \[\mu(t) = e^{ \int a(t)\,dt}\text{,}\] which makes the solution look a little bit less daunting: \[y(t) = \frac{1}{\mu(t)} \int b(t)\mu(t) \,dt + \frac{c}{\mu(t)} \text{.}\] Figure out what \(\mu'(t)\) is as a first step.]
Suppose that \(y_1(t)\) is a solution to the differential equation \(y' + a(t)y = 0\) and that \(y_2(t)\) is a solution to \(y' + a(t)y = b(t)\). Check that \(y_1(t) + y_2(t)\) is also a solution to \(y' + a(t)y = b(t)\). This is an important feature of linear equations: solutions can be built up from smaller pieces in some circumstances.
As noted, the previous problem heavily relies on the fact that the differential equation \(y' + a(t)y = b(t)\) is a linear equation. Consider the differential equation \[(y(t))^2 \cdot y''(t) + (y'(t))^3 = 0\text{.}\] Show that both \(y_1(t) = 1\) and \(y_2(t) = e^{-t}\) are solutions, but nonetheless \(1 + e^{-t}\) is not a solution. (Of course, this equation is wildly nonlinear.)
The recipe given for first-order differential equations doesn’t really require us to find antiderivatives at all the intermediate steps, which is good news because there are not many functions \(f(t)\) for which \(\int \exp\bigl(\int f(t)\,dt\bigr) \,dt\) can be calculated. Find a general solution to the differential equation \[e^t \,y'(t) - 5t^2 \,y(t) = \log(t+3) \text{.}\] Your answer will involve indefinite integrals that you shouldn’t be able to evaluate.
Some higher-order differential equations are actually first-order linear if you look at them in the right way. Use the substitutions suggested to solve the following equations. (The solution to the linear equation will be an explicit equation which then can be solved. You will have multiple constants.)
\(y'' + 5y' = t\); use \(w = y'\)
\(y''' + \frac{2}{t} y'' = 6\); use \(v = y'\) and \(w = v'\) and rewrite in terms of \(w\)
\(y' + (\alpha-1)y = 4\)
Answer the following for the above differential equation:
Find a general solution.
For what values of \(\alpha\) does a solution exist?
For what values of \(\alpha\) will the solution tend to infinity as \(t\rightarrow\infty\)?
If \(\alpha=1\) why do we not get \(y=4t\)?
What is \[x(t)=\frac{-1}{3}+3e^{3t}\] a solution to? What is the integrating factor?