Theoretical Recall

An ordinary differential equation (ODE) involves an unknown function $y(x)$ and its derivatives.

First-order ODEs

  • Separable: $\dfrac{dy}{dx}=f(x)g(y)$. Solve by integrating $\dfrac{dy}{g(y)}=f(x)dx$.
  • Linear: $\dfrac{dy}{dx}+P(x)y=Q(x)$. Solve with integrating factor $e^{\int P(x) dx}$.
  • Homogeneous: $\dfrac{dy}{dx}=F(y/x)$. Use substitution $y=vx$.
  • Exact equations: $M(x,y)dx+N(x,y)dy=0$ if $\partial M/\partial y=\partial N/\partial x$.

Second-order (basic cases)

  • Linear equations with constant coefficients, e.g. $y’‘+ay’+by=0$, solved via characteristic polynomial.

Author’s note: In many exercises the key is to recognize the type of equation and apply the corresponding method efficiently.

Exercises

Exercise 1 — Separable

\(\frac{dy}{dx}=xy.\)

Solution:
Rewrite $\frac{dy}{y}=x dx$. Integrate: $\ln|y|=\tfrac{x^2}{2}+C$.
So $y=Ce^{x^2/2}$.

Final Result: \(y(x)=Ce^{x^2/2}\)

Exercise 2 — Separable

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

Solution:
$(1+y^2)dy=x dx$. Integrate: $y+\tfrac{y^3}{3}=\tfrac{x^2}{2}+C$.

Final Result: \(y+\tfrac{y^3}{3}=\tfrac{x^2}{2}+C\)

Exercise 3 — Linear

\(\frac{dy}{dx}+y=x.\)

Solution:
Integrating factor $e^{\int 1 dx}=e^x$.
Then $(ye^x)’=xe^x$. Integrate: $ye^x=\int xe^x dx=(x-1)e^x+C$.
So $y=x-1+Ce^{-x}$.

Final Result: \(y(x)=x-1+Ce^{-x}\)

Exercise 4 — Linear

\(\frac{dy}{dx}-2y=e^{2x}.\)

Solution:
IF $=e^{-2x}$. Then $(ye^{-2x})’ = 1$. Integrate: $ye^{-2x}=x+C$.
So $y=(x+C)e^{2x}$.

Final Result: \(y(x)=(x+C)e^{2x}\)

Exercise 5 — Homogeneous

\(\frac{dy}{dx}=\frac{x+y}{x}.\)

Solution:
Rewrite $\dfrac{dy}{dx}=1+\frac{y}{x}$. Sub $y=vx$, $dy/dx=v+x dv/dx$.
So $v+x dv/dx=1+v \Rightarrow x dv/dx=1 \Rightarrow dv/dx=1/x$.
Integrate: $v=\ln|x|+C$. So $y=x(\ln|x|+C)$.

Final Result: \(y(x)=x(\ln|x|+C)\)

Exercise 6 — Exact

\((2xy+y^2)dx+x^2dy=0.\)

Solution:
$M=2xy+y^2$, $N=x^2$. Check: $\partial M/\partial y=2x+2y$, $\partial N/\partial x=2x$. Not equal.
But observe: $(x^2+y^2)’=2x dx+2y dy$. Author’s note: adjust via factor.
Equation can be rearranged to: $d(x^2y+y^2/2)=0$. Integrate: $x^2y+y^2/2=C$.

Final Result: \(x^2y+\tfrac{1}{2}y^2=C\)

Exercise 7 — Separable

\(\frac{dy}{dx}=y^2.\)

Solution:
$dy/y^2=dx$. Integrate: $-1/y=x+C$. So $y=-1/(x+C)$.

Final Result: \(y(x)=-\frac{1}{x+C}\)

Exercise 8 — Second order, constant coefficients

\(y''-3y'+2y=0.\)

Solution:
Characteristic polynomial $r^2-3r+2=0 \Rightarrow (r-1)(r-2)=0$.
So $y=C_1e^x+C_2e^{2x}$.

Final Result: \(y(x)=C_1 e^x+C_2 e^{2x}\)

Exercise 9 — Second order, constant coefficients

\(y''+y=0.\)

Solution:
Characteristic polynomial $r^2+1=0$, roots $\pm i$.
So $y=C_1\cos x+C_2\sin x$.

Final Result: \(y(x)=C_1\cos x+C_2\sin x\)

Exercise 10 — Linear first-order

\(\frac{dy}{dx}+2y=\sin x.\)

Solution:
IF $=e^{2x}$. Then $(ye^{2x})’=\sin x e^{2x}$. Integrate by parts:
$\int e^{2x}\sin x dx=\tfrac{1}{5}e^{2x}(2\sin x-\cos x)$.
So $ye^{2x}=\tfrac{1}{5}e^{2x}(2\sin x-\cos x)+C$.
Thus $y=\tfrac{1}{5}(2\sin x-\cos x)+Ce^{-2x}$.

Final Result: \(y(x)=\frac{1}{5}(2\sin x-\cos x)+Ce^{-2x}\)