Theoretical Recall
A Cauchy problem is an ODE together with initial conditions:
\[\begin{cases} y'(x) = F(x,y), \\ y(x_0) = y_0. \end{cases}\]For higher order: \(\begin{cases} y''+ay'+by=0, \\ y(x_0)=y_0,\; y'(x_0)=y_1. \end{cases}\)
Theorem (Picard–Lindelöf, simplified):
If $F(x,y)$ is continuous and satisfies a Lipschitz condition in $y$, then the Cauchy problem has a unique local solution.
Author’s note: Solving consists of two steps: find the general solution of the ODE, then use the initial conditions to fix the constants.
Exercises
Exercise 1
\(\begin{cases} y' = y, \\ y(0)=1. \end{cases}\)
Solution:
General solution: $y=Ce^x$. Condition: $1=C e^0 \Rightarrow C=1$.
Final Result: \(y(x)=e^x\)
Exercise 2
\(\begin{cases} y' = -2y, \\ y(0)=3. \end{cases}\)
Solution:
General: $y=Ce^{-2x}$. Condition: $3=C$.
Final Result: \(y(x)=3e^{-2x}\)
Exercise 3
\(\begin{cases} y' = x, \\ y(0)=0. \end{cases}\)
Solution:
Integrate: $y=\tfrac{x^2}{2}+C$. Condition: $0=C$.
Final Result: \(y(x)=\tfrac{x^2}{2}\)
Exercise 4
\(\begin{cases} y''-y=0, \\ y(0)=1,\; y'(0)=0. \end{cases}\)
Solution:
Characteristic: $r^2-1=0$, roots $\pm1$. General $y=C_1e^x+C_2e^{-x}$.
Conditions: $y(0)=C_1+C_2=1$, $y’(x)=C_1e^x-C_2e^{-x}$, so $y’(0)=C_1-C_2=0$.
Thus $C_1=C_2=1/2$.
Final Result: \(y(x)=\cosh x\)
Exercise 5
\(\begin{cases} y''+y=0, \\ y(0)=0,\; y'(0)=1. \end{cases}\)
Solution:
General $y=C_1\cos x+C_2\sin x$. Conditions: $y(0)=C_1=0$, $y’(x)=-C_1\sin x+C_2\cos x$, so $y’(0)=C_2=1$.
Final Result: \(y(x)=\sin x\)
Exercise 6
\(\begin{cases} y' - y = e^x, \\ y(0)=0. \end{cases}\)
Solution:
Linear ODE. IF $=e^{-x}$. Then $(ye^{-x})’ =1$. Integrate: $ye^{-x}=x+C$. So $y=(x+C)e^x$.
Condition: $0=(0+C)e^0 \Rightarrow C=0$.
Final Result: \(y(x)=x e^x\)
Exercise 7
\(\begin{cases} y''+4y=0, \\ y(0)=2,\; y'(0)=0. \end{cases}\)
Solution:
Characteristic: $r^2+4=0$, roots $\pm2i$. General $y=C_1\cos(2x)+C_2\sin(2x)$.
Conditions: $y(0)=C_1=2$, $y’(x)=-2C_1\sin(2x)+2C_2\cos(2x)$, $y’(0)=2C_2=0\Rightarrow C_2=0$.
Final Result: \(y(x)=2\cos(2x)\)
Exercise 8
\(\begin{cases} y''-3y'+2y=0, \\ y(0)=1,\; y'(0)=1. \end{cases}\)
Solution:
Characteristic $(r-1)(r-2)=0$, roots $1,2$. General $y=C_1e^x+C_2e^{2x}$.
Conditions: $y(0)=C_1+C_2=1$, $y’(x)=C_1e^x+2C_2e^{2x}$, so $y’(0)=C_1+2C_2=1$.
Subtract: $(C_1+2C_2)-(C_1+C_2)=C_2=0$. So $C_1=1$.
Final Result: \(y(x)=e^x\)
Exercise 9
\(\begin{cases} y''+y' = 0, \\ y(0)=0,\; y'(0)=1. \end{cases}\)
Solution:
Characteristic $r^2+r=0\Rightarrow r(r+1)=0$. Roots $0,-1$.
General $y=C_1+C_2e^{-x}$.
Derivative: $y’=-C_2e^{-x}$.
Conditions: $y(0)=C_1+C_2=0$, $y’(0)=-C_2=1\Rightarrow C_2=-1$, $C_1=1$.
Final Result: \(y(x)=1-e^{-x}\)
Exercise 10
\(\begin{cases} y' = y\cos x, \\ y(0)=1. \end{cases}\)
Solution:
Equation separable: $dy/y=\cos x dx$. Integrate: $\ln y=\sin x+C$.
So $y=Ce^{\sin x}$. Condition $y(0)=1 \Rightarrow C=1$.
Final Result: \(y(x)=e^{\sin x}\)