MATH 308 Lecture 3

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Lecture Notes

Linear

Exercise 1

$\begin{array}{rl}{y}^{\prime }(x)& ={\mathrm{e}}^x-\sin x\\ y(x)& ={\mathrm{e}}^x+cosx+C\end{array}$

"easy case" differential equation: solve for derivative and calculate the antiderivative.

Exercise 2.2

$x^2{y}^{\prime }(x)+2xy(x)={\mathrm{e}}^x-\sin x$

Let $g(x)=x^2$, then we have $g(x)y^{\prime }(x)+g^{\prime }(x)y(x)={\mathrm{e}}^x-\sin x$

Note the product rule on the LHS: now we have

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(x^{2}y\right)& ={\mathrm{e}}^x-\sin x\\ x^{2}y& ={\mathrm{e}}^x+\cos x+C\\ y& =\frac{{\mathrm{e}}^x}{x^2}+\frac{\cos x}{x^2}+\frac{C}{x^2}\end{array}$

Exercise 2.2

Note: $\frac{\mathrm{d}}{\mathrm{d}x}({\mathrm{e}}^{x}y(x))={\mathrm{e}}^x(y^{\prime }(x)+y(x))$

$\begin{array}{rl}{y}^{\prime }(x)+y(x)& =1+e^x\\ {\mathrm{e}}^x{y}^{\prime }(x)+{\mathrm{e}}^{x}y(x)& ={\mathrm{e}}^x+{\mathrm{e}}^{2x}\\ y(x)& =1+\frac{{\mathrm{e}}^x}{2}+\frac{C}{{\mathrm{e}}^x}\end{array}$

Exercise 3

$\begin{array}{rl}{y}^{\prime }(x)-y(x)\tan x& =\sin x\cos x\\ \mu (x)& =\exp ((\int -\frac{\sin x}{\cos x}\mathrm{d}x)=\cos x\\ \cos x{y}^{\prime }(x)-\sin xy(x)& ={\cos }^{2}x\sin x\\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos xy(x))& ={\cos }^{2}x\sin x\\ \cos xy(x)& =\frac{1}{3}{\cos }^{3}x+C\\ y(x)& =\frac{1}{3}{\cos }^{2}x+C\sec x\end{array}$

Integrating Factor

For an ODE of the form

$y^{\prime }(x)+P(x)y(x)=g(x)$

The integrating factor is defined as

$\mu (x)=\exp (\int P(x)\mathrm{d}x)$

such that the following ODE is solvable by the product rule:

$\begin{array}{rl}\mu (x)(y^{\prime }(x)+P(x)y(x))& =\mu (x)g(x)\\ \frac{\mathrm{d}}{\mathrm{d}x}(\mu (x)y(x))& =\mu (x)g(x)\\ \mu (x)y(x)& =\int \mu (x)g(x)\mathrm{d}x\\ y(x)& =\frac{1}{\mu (x)}\int \mu (x)g(x)\mathrm{d}x\end{array}$

In Maple, type

> intfactor(ODE);

Exercise 4

Solve the ODE $x{y}^{\prime }-2y=\sqrt{x}$ given the initial condition $y(1)=0$

Rewrite in form $y^{\prime }-\frac{2}{x}y=\frac{1}{\sqrt{x}}$ to find the integrating factor $\mu (x)=x^{-2}$.

$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{y}{x^2}\right)& =x^{-\frac{5}{2}}\\ \frac{y}{x^2}& =\frac{2}{3}{x}^{-\frac{3}{2}}+C\\ \end{array}$

At this point, we plug in our initial conditions to solve for $C$ before finding $y$.

$\begin{array}{rl}\frac{0}{1}& =-\frac{2}{3}+C\\ C& =\frac{2}{3}\end{array}$

Now we can solve the ODE for $y$.

$y=\frac{2}{3}\sqrt{x}+\frac{2}{3}{x}^2$

Exercise 5

$\begin{array}{rl}{y}^{\prime }+\frac{t+1}{t}y& =1\\ \mu (t)& =\exp (\int 1+t^{-1}\mathrm{d}x)={\mathrm{e}}^{t+\ln t}\\ \frac{\mathrm{d}}{\mathrm{d}t}({\mathrm{e}}^{t+\ln t}y(t))& ={\mathrm{e}}^{t+\ln t}=t{\mathrm{e}}^t\\ t{\mathrm{e}}^{t}y(t)& =t{\mathrm{e}}^t-{\mathrm{e}}^t+C\\ y(t)& =1-t^{-1}+\frac{C}{t{\mathrm{e}}^t}\end{array}$

Given the inital condition $y(\ln 2)=1$ (assuming $t>0$), the particular solution is

$y(t)=1-t^{-1}+\frac{2}{t{\mathrm{e}}^t}$