Consider a more general case of first order linear differential equation: dy/dt + U(t)y =W(t): when U(t) and W(t) represent a variable coefficient and variable term, respectively. i. Find the general solution of the equation dy/dt + 3t^2 y= 0. ii. Find the general solution of the equation dy/dt + 2ty= t iii. Find the equation dy/dt +4 =4t?

Question

Answer

Solution:

To find the general solution of the given first-order linear differential equations, we will use the method of integrating factors.

General Steps:
1. Write the given differential equation in the standard form: dy/dt + P(t)y = Q(t).
2. Identify the coefficient P(t) and the term Q(t).
3. Find the integrating factor, denoted by μ(t), which is defined as μ(t) = e^(∫P(t)dt).
4. Multiply both sides of the equation by the integrating factor μ(t).
5. Rewrite the left side of the equation as the derivative of the product y(t)μ(t).
6. Integrate both sides of the equation.
7. Solve for y(t) to obtain the general solution.

Let's solve each part of the question step by step.

i. dy/dt + 3t^2y = 0:
1. The given differential equation is already in the standard form.
2. P(t) = 3t^2 and Q(t) = 0.
3. The integrating factor μ(t) = e^(∫3t^2dt) = e^t^3.
4. Multiplying both sides of the equation by e^t^3, we get e^t^3dy/dt + 3t^2e^t^3y = 0.
5. The left side of the equation can be rewritten as the derivative of the product y(t)e^t^3: d/dt(y(t)e^t^3) = 0.
6. Integrating both sides of the equation, we obtain y(t)e^t^3 = C, where C is the constant of integration.
7. Solving for y(t), we have y(t) = Ce^(-t^3).

Therefore, the general solution of the equation dy/dt + 3t^2y = 0 is y(t) = Ce^(-t^3), where C is any constant.

ii. dy/dt + 2ty = t:
1. The given differential equation is already in the standard form.
2. P(t) = 2t and Q(t) = t.
3. The integrating factor μ(t) = e^(∫2tdt) = e^(t^2).
4. Multiplying both sides of the equation by e^(t^2), we get e^(t^2)dy/dt + 2te^(t^2)y = te^(t^2).
5. The left side of the equation can be rewritten as the derivative of the product y(t)e^(t^2): d/dt(y(t)e^(t^2)) = te^(t^2).
6. Integrating both sides of the equation, we obtain y(t)e^(t^2) = (1/2)e^(t^2) + C, where C is the constant of integration.
7. Solving for y(t), we have y(t) = (1/2) + Ce^(-t^2).

Therefore, the general solution of the equation dy/dt + 2ty = t is y(t) = (1/2) + Ce^(-t^2), where C is any constant.

iii. dy/dt =

Similar Economics Doubts

Solve equation of dy/dt +4= 4t?

Solve the equation dy/dt+4y=0 with initial condition y(0) = 0?

Solve the first-order difference equation yt+1-5yt=1 given (yo=7/4)?

What is the defferential equation, homogeneous equation and non homogeneous equation and give examples?

Derivative of slutsky equation?

Top Courses for Economics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams