[EXAMPLES] ODE fundamentals

The relation

y+\tan^{-1}y=\frac{x^2}{2}

defines a solution of

y'=\frac{x(1+y^2)}{2+y^2}.

Differentiating implicitly gives

y'+\frac{y'}{1+y^2}=x.

Collecting the terms containing $y'$ ,

y'\frac{2+y^2}{1+y^2}=x,

y'=\frac{x(1+y^2)}{2+y^2}.

[routine]

For a first-order equation,

y'=\sin(3x)

gives

y=-\frac13\cos(3x)+C.

For a third-order equation,

y'''=0

gives

y=Ax^2+Bx+C.

For

y''=x+5,

integrating twice gives

y=\frac{x^3}{6}+\frac{5x^2}{2}+C_1x+C_2.

[routine]

For the IVP

y'=\sin(3x),\qquad y(0)=4,

the general solution gives

-\frac13+C=4.

Thus

y=4+\frac{1-\cos(3x)}{3}.

For the BVP

y''+6y'+9y=0,\qquad y(0)=1,\qquad y(2)=0,

the repeated-root solution is

y=(A+Bx)e^{-3x}.

The first condition gives $A=1$ . The second gives

(1+2B)e^{-6}=0,

so $B=-\frac12$ . The BVP solution is therefore

y=\left(1-\frac{x}{2}\right)e^{-3x}.

[exam]

The instruction "solve the IVP" can have different outcomes:

$y'=x$ , $y(0)=1$ has the unique solution $y=1+x^2/2$ .
$y'=3y^{2/3}$ , $y(0)=0$ has at least the two solutions $y=0$ and $y=x^3$ , so uniqueness can fail. The Lipschitz condition explains why.
$xy'=y-1$ , $y(0)=1$ has the family $y=1+Cx$ ; the coefficient of $y'$ vanishes at the initial point, so the usual normal form is singular there.
$|y'|+|y|=0$ , $y(0)=1$ has no solution, because the equation forces $y=0$ everywhere while the initial value requires $y(0)=1$ .

[challenge]

Examples from ODE fundamentals