This is the review note for Math285 Midterm (Spring 2025).

First Order ODE

Theorem

To solve $y’ = a(t)y + b(t)$:

1. Find $A(t)$: Calculate the integral of $a(t)$: $A(t) = \int a(t) \, dt$. You can ignore the constant of integration for this step.

2. Calculate $e^{A(t)}$ and $e^{-A(t)}$: Compute the exponential of $A(t)$ and its negative.

3. Find the Integral Part: Calculate the integral of $b(t) \cdot e^{-A(t)}$: $\int b(t)e^{-A(t)} \, dt$. Let’s call the result of this integral $I(t)$.

4. Particular Solution: A particular solution is given by $y_p(t) = e^{A(t)} \cdot I(t)$.

5. General Solution: The general solution is $y(t) = c \cdot e^{A(t)} + y_p(t)$, where $c$ is an arbitrary constant.

sample

1 For the solution $y(t)$ of the IVP $y’ = (y/t) - 1$, $y(1) = \ln 2$, the value $y(2)$ is equal to:

• 0
• 1
• 2
• $\ln 2$
• $2\ln 2$

2. For the solution $y(t)$ of the IVP $y’ = \frac{ty+1}{t^2+1}$, $y(0) = 2$, the value $y(1)$ is equal to:

• $\sqrt{2}$
• 2
• $1+\sqrt{2}$
• 3
• $1+2\sqrt{2}$

3. For the solution $y(t)$ of the IVP $y’ = \frac{2y+1}{t}$, $y(1) = 2$, the value $y(2)$ is equal to:

• 11/2
• 13/2
• 15/2
• 17/2
• 19/2

Separatble ODEs

Theorem

Given a separable ODE of the form: $\frac{dy}{dt} = f(t)g(y)$

1. Separate: $\frac{dy}{g(y)} = f(t) dt$

2. Integrate: $\int \frac{1}{g(y)} dy = \int f(t) dt$

   Let $G(y) = \int \frac{1}{g(y)} dy$ and $F(t) = \int f(t) dt$

   Then, the equation becomes: $G(y) = F(t) + C$ (where $C$ is the integration constant)

3. Apply Initial Condition $y(t_0) = y_0$: $G(y_0) = F(t_0) + C$

4. Solve for $C$: $C = G(y_0) - F(t_0)$

5. Specific Solution (Implicit Form): $G(y) = F(t) + G(y_0) - F(t_0)$

6. Solve for $y$ (Explicit Form - if possible): If possible, solve $G(y) = F(t) + C$ (or $G(y) = F(t) + G(y_0) - F(t_0)$) for $y$ to get $y = h(t, C)$ or $y = h(t, t_0, y_0)$.

7. Evaluate $y(t_1)$: Substitute $t = t_1$ into the explicit solution $y = h(t, C)$ (or implicit solution and solve for $y$) to find $y(t_1)$.

Sample

1. For the solution $y(t)$ of the IVP $y’ = y\ln t$, $y(1) = 1$ the value $y(e)$ is equal to

   • $e^{-2}$
   • $e^{-1}$
   • 1
   • $e$
   • $e^{2}$

2. For the solution $y(t)$ of the IVP $y’ = y^2e^{-t}$, $y(0) = -1$ the value $y(-1)$ is equal to

   • $e$
   • $1/(e-2)$
   • $e+2$
   • $1/(e+2)$
   • $e-2$

3. For the solution $y(t)$ of the IVP $y’ = e^{t-2y}$, $y(0) = 0$ the value $y(1)$ is contained in

   • $[0, \frac{1}{2}]$
   • $[\frac{1}{2}, 1]$
   • $[1, \frac{3}{2}]$
   • $[\frac{3}{2}, 2]$
   • $[2, \infty)$

4. For the solution $y(t)$ of the IVP $y’ = (y^2-3)/(ty)$, $y(1) = 2$ the value $y(2)$ is equal to

   • $\sqrt{6}$
   • $\sqrt{7}$
   • $\sqrt{8}$
   • $3$
   • $\sqrt{10}$

5. For the solution $y(t)$ of the IVP $y’ = -t(y^2+1)$, $y(0) = 1$ the value $y(1)$ is contained in

   • $[0, \frac{1}{2}]$
   • $[\frac{1}{2}, 1]$
   • $[1, \frac{3}{2}]$
   • $[\frac{3}{2}, 2]$
   • $[2, \infty)$