Solutions for differential equations

First, apply separation of variables and integrate both sides:

\begin{align*} \frac{dy}{dt} = \frac{2y(t) - 1}{t+3} &\Rightarrow \frac{dy}{2y(t) - 1} = \frac{dt}{t+3}\\ &\Rightarrow \int \frac{dy}{2y(t) - 1} = \int \frac{dt}{t+3}\\ &\Rightarrow \frac{1}{2}\ln|2 y(t) - 1| = \ln|t+3| + C \end{align*}

(1)

To determine the particular solution to the initial-value problem, we now solve for the constant $C$ :

\begin{align*} \frac{1}{2}\ln|2 y(t) - 1| = \ln|t+3| + C &\Rightarrow \frac{1}{2}\ln|2 (-1) - 1| = \ln|0+3| + C\\ &\Rightarrow \frac{1}{2}\ln|-3| = \ln|3| + C\\ &\Rightarrow \frac{1}{2}\ln(3) = \ln(3) + C\\ &\Rightarrow - \frac{1}{2}\ln(3) = C \end{align*}

(2)

We now substitute in our value of $C$ to obtain the explicit solution:

\begin{align*} \frac{1}{2}\ln|2 y(t) - 1| &= \ln|t+3| - \frac{1}{2}\ln(3)\\ &\Rightarrow \ln|2 y(t) - 1| = 2\ln|t+3| - \ln(3)\\ &\Rightarrow \ln|2y(t) - 1| = \ln|(t+3)^2| - \ln(3)\\ &\Rightarrow \ln|2y(t) -1| = \ln|\frac{(t+3)^2}{3}|\\ &\Rightarrow 2y(t) - 1 = \frac{(t+3)^2}{3}\\ &\Rightarrow \boxed{y(t) = \frac{(t+3)^2}{6} - \frac{1}{2}} \end{align*}

(3)

Solution to Exercise 2

For this problem you will need to use the method of integrating factors. First, rearrange and identify the functions $p(t)$ and $q(t)$ in the general form $y'(t) + p(t) y(t) = q(t)$ :

\begin{align*} (1+t^2)y'(t) - 2ty(t) = t &\Rightarrow y'(t) - \frac{2t}{1 + t^2} y(t) = \frac{t}{1+t^2} \end{align*}

(4)

p(t) = - \frac{2t}{1 + t^2}, q(t) = \frac{t}{1+t^2}

Now, solve for the integrating factor $\mu$ via integration:

\begin{align*} \mu &= \exp(\int p(t)dt)\\ &= \exp(\int - \frac{2t}{1 + t^2}dt)\\ &= \exp(- \ln(1 + t^2))\\ &= \exp( \ln(\frac{1}{1+t^2}))\\ &= \frac{1}{1+t^2} \end{align*}

(5)

We can now use our obtained integrating factor to solve for $y(t)$ :

\begin{align*} y(t) &= \frac{1}{\mu}\left[\int \mu q(t)dt + C\right]\\ &= (1 + t^2)\left[\int \frac{1}{(1+t^2)^2} dt + C\right]\\ &= (1 + t^2)\left[\frac{1}{2(1+t^2)}\ln|(1+t^2)^2| + C\right]\\ &= \frac{1}{2} \ln|(1 + t^2)| + C(1+t^2) \end{align*}

(6)

\begin{align*} y(t) = \frac{1}{2} \ln|(1 + t^2)| + C(1+t^2) &\Rightarrow 0 = \frac{1}{2} \ln|1| + C\\ &\Rightarrow C = -\frac{1}{2} \ln(1)\\ &\Rightarrow C = 0 \end{align*}

(7)

After applying the initial condition we therefore obtain the explicit solution:

\boxed{y(t) = \frac{1}{2} \ln|(1 + t^2)|}

(8)

Solution to Exercise 3

This is a special case of a nonlinear differential equation that can be solved analytically. To start, let’s algebraically rearrange to get the following form:

\begin{align*} y' - \frac{2y}{x} = -x^2y^2 &\Rightarrow \frac{y'}{y^2} - \frac{2}{xy} = -x^2 \end{align*}

(9)

We’ll now use a change of variables of $u=\frac{1}{y}, u' = \frac{-y'}{y^2}$

\begin{align*} \frac{y'}{y^2} - \frac{2}{xy} = -x^2 &\Rightarrow -u' - \frac{2}{x}u = -x^2\\ &\Rightarrow u' + \frac{2}{x}u = x^2 \end{align*}

(10)

This has turned a nonlinear differential equation into a linear differential equation, which we can now solve by the method of integrating factors: