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Magazine Chapter 1: Problem 20
\(d y / d x+2 y=x^{2}, y(0)=1, y=\frac{1}{4}-\frac{1}{2} x+\) \(\frac{1}{2} x^{2}+A e^{-2 x}\)
Short Answer
Expert verified
The final solution is \(y = e^{-2x} + \frac{1}{2}x^2\).
Step by step solution
01
Identify the given differential equation
The given differential equation is \(\frac{dy}{dx} + 2y = x^2\).
02
Identify the complementary function
Solve the homogeneous part \(\frac{dy}{dx} + 2y = 0\). The complementary function is \(y_c = Ae^{-2x}\), where \(A\) is a constant.
03
Find the particular solution
Assume a particular solution for \(x^2\). Given that the non-homogeneous part is a polynomial \(x^2\), try \(y_p = ax^2 + bx + c\) as a particular solution.
04
Determine constants of the particular solution
Differentiate the assumed particular solution: \(y_p = ax^2 + bx + c.\) \(\frac{dy_p}{dx} = 2ax + b.\) Substitute \(y_p\) and \(\frac{dy_p}{dx}\) back into the differential equation: \(2ax + b + 2(ax^2 + bx + c) = x^2.\) Equate coefficients of like powers of \(x\): \(2a = 1,\) \(2b = 0,\) \(2c = 0.\) Thus, \(a = \frac{1}{2},\ b = 0,\ and\ c = 0.\)
05
Write the general solution
Combine the complementary function and particular solution: \(y = y_c + y_p = Ae^{-2x} + \frac{1}{2}x^2\).
06
Apply the initial condition
Given \(y(0) = 1\), plug \(x = 0\) into the general solution: \(1 = A e^{-2(0)} + \frac{1}{2}(0)^2 + 0\). Simplifies to \(1 = A\). Thus, \(A = 1\).
07
State the final solution
Combine all terms to give the final answer: \(y = e^{-2x} + \frac{1}{2}x^2\). Match to given form \(y = \frac{1}{4} - \frac{1}{2}x + \frac{1}{2}x^2 + Ae^{-2x}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Solutions
A homogeneous solution to a differential equation involves solving the equation with the right-hand side set to zero. For instance, the given equation is \(\frac{dy}{dx} + 2y = x^2\). The homogeneous form is \(\frac{dy}{dx} + 2y = 0\).
To solve this, we assume a solution of the form \y = Ce^{kx}\, where C is a constant and k is a parameter to be determined. By substituting this into the homogeneous equation and solving for k, we find our complementary function.
To solve this, we assume a solution of the form \y = Ce^{kx}\, where C is a constant and k is a parameter to be determined. By substituting this into the homogeneous equation and solving for k, we find our complementary function.
- Homogeneous equation: Set the right-hand side to zero.
- Test a solution of form \y = Ce^{kx}\.
- Find the parameter k by solving the resulting algebraic equation.
Non-Homogeneous Differential Equations
A non-homogeneous differential equation includes a term that does not depend on the function and its derivatives alone. Here, we have the equation \(\frac{dy}{dx} + 2y = x^2\). This non-homogeneous term is \x^2\.
To solve these equations:
To solve these equations:
- Solve the homogeneous part to get the complementary function.
- Find the particular solution for the non-homogeneous term.
- Add the complementary and particular solutions.
Particular Solution
To find a particular solution to a non-homogeneous differential equation, we assume a form similar to the non-homogeneous term. For the equation \(\frac{dy}{dx} + 2y = x^2\), we assume \y_p = ax^2 + bx + c\.
- Differentiate the assumed solution.
- Substitute back into the original equation.
- Solve for the unknown coefficients by equating like terms.
Initial Condition
An initial condition is specific information about the value of the solution at a given point, which helps in finding the particular constant. Given \ y(0) = 1\, we substitute \x = 0\ into the general solution to find our constant.
- Plug in the given value of x.
- Set the function equal to the initial condition.
- Solve for the constant.
Complementary Function
The complementary function is the solution to the homogeneous part of the differential equation. For our original equation, we isolate the homogeneous portion: \(\frac{dy}{dx} + 2y = 0\).
- Assume a solution of the form \y = Ce^{kx}\.
- Substitute and solve for k.
- Derive the general form of the complementary function.
