Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 27
Solve the following differential equations with the given initial conditions. $$ y^{\prime}=5 t y-2 t, y(0)=1 $$
Short Answer
Expert verified
The solution is \( y(t) = \frac{2}{5} + \frac{3}{5} e^{\frac{5t^2}{2}} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is first-order linear: \[ y^{\rprime} = 5ty - 2t \] with the initial condition \( y(0) = 1 \).
02
Rewrite in Standard Form
Rewrite the equation in the standard form \( y^{\rprime} + P(t)y = Q(t) \). Given: \[ y^{\rprime} - 5ty = -2t \] Where, \(P(t) = -5t\) and \(Q(t) = -2t\).
03
Find the Integrating Factor
The integrating factor \( \rho(t) \) is given by: \[ \rho(t) = e^{\int P(t) \, dt} = e^{\int -5t \, dt} = e^{-\frac{5t^2}{2}} \]
04
Multiply the Differential Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor: \[ e^{-\frac{5t^2}{2}}y^{\rprime} - 5te^{-\frac{5t^2}{2}}y = e^{-\frac{5t^2}{2}}(-2t) \] Simplifying, we get: \[ \left( e^{-\frac{5t^2}{2}} y \right)^{\rprime} = -2t e^{-\frac{5t^2}{2}} \]
05
Integrate Both Sides
Integrate both sides with respect to \( t \): \[ e^{-\frac{5t^2}{2}} y = \int -2t e^{-\frac{5t^2}{2}} \, dt \] Let \( u = -\frac{5t^2}{2} \), then \( du = -5t \, dt \): \[ \int -2t e^{-\frac{5t^2}{2}} \, dt = \int \left( \frac{2}{5} \right) e^{u} \, du \] This evaluates to: \[ \left( \frac{2}{5} \right) e^{u} = \left( \frac{2}{5} \right) e^{-\frac{5t^2}{2}} \]
06
Solve for y and Apply Initial Condition
Thus, we have: \[ e^{-\frac{5t^2}{2}} y = \left( \frac{2}{5} \right) e^{-\frac{5t^2}{2}} + C \] Multiply both sides by \( e^{\frac{5t^2}{2}} \): \[ y = \frac{2}{5} + Ce^{\frac{5t^2}{2}} \] Using the initial condition \( y(0) = 1 \): \[ 1 = \frac{2}{5} + C \] Thus, \( C = \frac{3}{5} \).
07
Write the Final Solution
Substitute the value of \( C \) back into the solution: \[ y(t) = \frac{2}{5} + \frac{3}{5} e^{\frac{5t^2}{2}} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are equations that involve a function and its derivatives. In our problem, we have a first-order linear differential equation, represented by \( y^{\rprime} = 5ty - 2t \), where \( y^{\rprime} \) represents the derivative of \( y \) with respect to \( t \). These equations are prominently used to model real-world phenomena like motion, growth, decay, and more.
- First-order means it involves the first derivative of the function.
- Linear indicates that both the function \( y \) and its derivative \( y^{\rprime} \) appear to the power of one.
Initial Conditions
Initial conditions specify the value of the function at a particular point. For instance, the initial condition in our problem is given as \( y(0) = 1 \).
These conditions are vital because:
These conditions are vital because:
- They help determine the specific solution to the differential equation from a family of possible solutions.
- Initial conditions provide constraints that make the solution unique to the problem context.
Integrating Factor
An integrating factor is a crucial tool used in solving linear differential equations. It's a function that, when multiplied with the original differential equation, allows it to be rewritten in an easier form.
For our equation, the integrating factor \( \rho(t) \) is given by: \[ \rho(t) = e^{\int P(t) dt} = e^{-\frac{5t^2}{2}} \]
The steps to use an integrating factor are:
For our equation, the integrating factor \( \rho(t) \) is given by: \[ \rho(t) = e^{\int P(t) dt} = e^{-\frac{5t^2}{2}} \]
The steps to use an integrating factor are:
- Identify \( P(t) \) from the standard form of the differential equation.
- Compute the integrating factor \( \rho(t) \).
- Multiply the entire differential equation by \( \rho(t) \).
Integration by Substitution
Integration by substitution is a method that simplifies the integration process by changing variables. For our problem, we make a substitution to handle the integration:
Let \( u = -\frac{5t^2}{2} \), then \( du = -5t \, dt \).
The original integral then becomes easier to handle:
\[ \int -2t e^{u} dt = \int \left( \frac{2}{5} \right) e^{u} du \]
This step effectively translates a complicated integral into a simpler one. Here's a breakdown of our substitution process:
Let \( u = -\frac{5t^2}{2} \), then \( du = -5t \, dt \).
The original integral then becomes easier to handle:
\[ \int -2t e^{u} dt = \int \left( \frac{2}{5} \right) e^{u} du \]
This step effectively translates a complicated integral into a simpler one. Here's a breakdown of our substitution process:
- Select an appropriate substitution \( u \) to simplify the integrand.
- Express \( dt \) in terms of \( du \).
- Rewrite the integral with the new variable \( u \).
