Math Problem Statement

Solve the differential equation for y(t) with y(0) = y_0. a) a·y' + p·y = 0, b) m·y' + b·y = 0, c) R·C·y' + y = 0

Solution

The image contains differential equations that need to be solved for y(t)y(t) with the initial condition y(0)=y0y(0) = y_0. Let's break down and solve each one:

Part a)

ay˙+py=0a \cdot \dot{y} + p \cdot y = 0

This is a first-order linear differential equation. Let's solve it step by step:

  1. Rewrite the equation: y˙=pay\dot{y} = -\frac{p}{a} \cdot y

  2. This can be solved using the separation of variables technique. Separate yy and tt as follows: 1ydy=padt\frac{1}{y} \, dy = -\frac{p}{a} \, dt

  3. Integrating both sides: 1ydy=padt\int \frac{1}{y} \, dy = \int -\frac{p}{a} \, dt lny=pat+C\ln |y| = -\frac{p}{a} t + C

  4. Solve for yy: y(t)=C1epaty(t) = C_1 e^{-\frac{p}{a} t}

  5. Apply the initial condition y(0)=y0y(0) = y_0: y0=C1e0=C1y_0 = C_1 e^0 = C_1 Therefore, C1=y0C_1 = y_0, and the final solution is: y(t)=y0epaty(t) = y_0 e^{-\frac{p}{a} t}

Part b)

my˙+by=0m \cdot \dot{y} + b \cdot y = 0

This is of the same form as part (a). Following the same steps:

  1. Rewrite: y˙=bmy\dot{y} = -\frac{b}{m} \cdot y

  2. Separate variables and integrate: 1ydy=bmdt\frac{1}{y} \, dy = -\frac{b}{m} \, dt lny=bmt+C\ln |y| = -\frac{b}{m} t + C y(t)=C2ebmty(t) = C_2 e^{-\frac{b}{m} t}

  3. Apply the initial condition y(0)=y0y(0) = y_0: y(t)=y0ebmty(t) = y_0 e^{-\frac{b}{m} t}

Part c)

RCy˙+y=0R \cdot C \cdot \dot{y} + y = 0

Again, this is a first-order linear differential equation. Follow the same method as in parts (a) and (b):

  1. Rewrite: y˙=1RCy\dot{y} = -\frac{1}{R \cdot C} \cdot y

  2. Separate variables and integrate: 1ydy=1RCdt\frac{1}{y} \, dy = -\frac{1}{R \cdot C} \, dt lny=1RCt+C\ln |y| = -\frac{1}{R \cdot C} t + C y(t)=C3e1RCty(t) = C_3 e^{-\frac{1}{R \cdot C} t}

  3. Apply the initial condition y(0)=y0y(0) = y_0: y(t)=y0e1RCty(t) = y_0 e^{-\frac{1}{R \cdot C} t}

Summary of Solutions:

  • (a) y(t)=y0epaty(t) = y_0 e^{-\frac{p}{a} t}
  • (b) y(t)=y0ebmty(t) = y_0 e^{-\frac{b}{m} t}
  • (c) y(t)=y0e1RCty(t) = y_0 e^{-\frac{1}{R \cdot C} t}

These are standard solutions for first-order linear differential equations with constant coefficients.

Would you like further details or explanations on any of these solutions?

Here are 5 follow-up questions to expand your understanding:

  1. How do you solve non-homogeneous first-order differential equations?
  2. What is the physical meaning of these differential equations in electrical circuits?
  3. How would you approach solving a second-order differential equation?
  4. What is the significance of the exponential term in the solutions?
  5. How does the method of integrating factors apply to solving linear differential equations?

Tip: When solving first-order differential equations, always try separation of variables first—it simplifies many common problems!

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Linear First-Order Differential Equations
Separation of Variables

Formulas

a·dy/dt + p·y = 0
m·dy/dt + b·y = 0
R·C·dy/dt + y = 0

Theorems

Solution of Linear First-Order Differential Equations
Exponential Decay

Suitable Grade Level

Undergraduate (First-Year Calculus or Differential Equations Course)