MATH 308 Lecture 3

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Lecture Notes

Linear

Exercise 1

${\begin{aligned}y'(x)&=\mathrm {e} ^{x}-\sin {x}\\y(x)&=\mathrm {e} ^{x}+cos{x}+C\end{aligned}}$

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

Exercise 2.2

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

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

Note the product rule on the LHS: now we have

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

Exercise 2.2

Note: ${\frac {\mathrm {d} }{\mathrm {d} x}}\left(\mathrm {e} ^{x}\,y(x)\right){=}\mathrm {e} ^{x}\left(y'(x)+y(x)\right)$

${\begin{aligned}y'(x)+y(x)&=1+e^{x}\\\mathrm {e} ^{x}\,y'(x)+\mathrm {e} ^{x}\,y(x)&=\mathrm {e} ^{x}+\mathrm {e} ^{2x}\\y(x)&=1+{\frac {\mathrm {e} ^{x}}{2}}+{\frac {C}{\mathrm {e} ^{x}}}\end{aligned}}$

Exercise 3

${\begin{aligned}y'(x)-y(x)\,\tan {x}&=\sin {x}\,\cos {x}\\\mu (x)&=\exp(\left(\int -{\frac {\sin {x}}{\cos {x}}}\,\mathrm {d} x\right)=\cos {x}\\\cos {x}\,y'(x)-\sin {x}\,y(x)&=\cos ^{2}{x}\,\sin {x}\\{\frac {\mathrm {d} }{\mathrm {d} x}}\left(\cos {x}\,y(x)\right)&=\cos ^{2}{x}\,\sin {x}\\\cos {x}\,y(x)&={\frac {1}{3}}\,\cos ^{3}{x}+C\\y(x)&={\frac {1}{3}}\,\cos ^{2}{x}+C\sec {x}\end{aligned}}$

Integrating Factor

For an ODE of the form

$y'(x)+P(x)\,y(x)=g(x)$

The integrating factor is defined as

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

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

${\begin{aligned}\mu (x)\left(y'(x)+P(x)\,y(x)\right)&=\mu (x)\,g(x)\\{\frac {\mathrm {d} }{\mathrm {d} x}}\left(\mu (x)\,y(x)\right)&=\mu (x)\,g(x)\\\mu (x)\,y(x)&=\int \mu (x)\,g(x)\,\mathrm {d} x\\y(x)&={\frac {1}{\mu (x)}}\,\int \mu (x)\,g(x)\,\mathrm {d} x\end{aligned}}$

In Maple, type

> intfactor(ODE);

Exercise 4

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

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

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

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

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

Now we can solve the ODE for $y$.

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

Exercise 5

${\begin{aligned}y'+{\frac {t+1}{t}}\,y&=1\\\mu (t)&=\exp {\left(\int 1+t^{-1}\,\mathrm {d} x\right)}=\mathrm {e} ^{t+\ln {t}}\\{\frac {\mathrm {d} }{\mathrm {d} t}}\left(\mathrm {e} ^{t+\ln {t}}\,y(t)\right)&=\mathrm {e} ^{t+\ln {t}}=t\,\mathrm {e} ^{t}\\t\,\mathrm {e} ^{t}\,y(t)&=t\,\mathrm {e} ^{t}-\mathrm {e} ^{t}+C\\y(t)&=1-t^{-1}+{\frac {C}{t\,\mathrm {e} ^{t}}}\end{aligned}}$

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

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