Find the general solution of this ODE: d²y dt² +13- +36y 4t² - 6+10e-2t dy dt = The solution will be of the form: y(t) = Cy₁(t) + Dy2(t) + yp(t) so use C and D as the arbitrary constants. y(t) =
Find the general solution of this ODE: d²y dt² +13- +36y 4t² - 6+10e-2t dy dt = The solution will be of the form: y(t) = Cy₁(t) + Dy2(t) + yp(t) so use C and D as the arbitrary constants. y(t) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Find the general solution of this ODE:
d2ydt2+13dydt+36y=4t2−6+10e−2td2ydt2+13dydt+36y=4t2-6+10e-2t
The solution will be of the form:
y(t)=Cy1(t)+Dy2(t)+yp(t)y(t)=Cy1(t)+Dy2(t)+yp(t)
so use CC and DD as the arbitrary constants.
![### Solving Second-Order Ordinary Differential Equations (ODEs)
**Problem Statement:**
Find the general solution of this ODE:
\[
\frac{d^2y}{dt^2} + 13\frac{dy}{dt} + 36y = 4t^2 - 6 + 10e^{-2t}
\]
**Solution:**
The solution will be of the form:
\[
y(t) = C y_1(t) + D y_2(t) + y_p(t)
\]
where \( y_1(t) \) and \( y_2(t) \) are the solutions to the homogeneous equation, and \( y_p(t) \) is a particular solution to the non-homogeneous equation.
So use \( C \) and \( D \) as the arbitrary constants.
The general solution for \( y(t) \) is:
\[
y(t) = \boxed{ \ \ }
\]
**Explanation:**
In these types of problems, we first solve the homogeneous differential equation:
\[
\frac{d^2y}{dt^2} + 13\frac{dy}{dt} + 36y = 0
\]
This will give us the complementary solution involving \( y_1(t) \) and \( y_2(t) \).
Then, we find a particular solution \( y_p(t) \) for the non-homogeneous part \( 4t^2 - 6 + 10e^{-2t} \).
Combining the complementary solution and the particular solution gives us the general solution.
**NOTE:**
In practice, solving each part requires additional steps, such as finding the roots of the characteristic equation for the complementary part and applying methods like undetermined coefficients or variation of parameters for the particular solution. This box is reserved for inputting the final general solution after completing those steps.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22e84192-4b16-4dc2-817c-afb771201e2b%2F9273d8f2-a41a-4a10-abda-185aa1827a2d%2F2io9n9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Second-Order Ordinary Differential Equations (ODEs)
**Problem Statement:**
Find the general solution of this ODE:
\[
\frac{d^2y}{dt^2} + 13\frac{dy}{dt} + 36y = 4t^2 - 6 + 10e^{-2t}
\]
**Solution:**
The solution will be of the form:
\[
y(t) = C y_1(t) + D y_2(t) + y_p(t)
\]
where \( y_1(t) \) and \( y_2(t) \) are the solutions to the homogeneous equation, and \( y_p(t) \) is a particular solution to the non-homogeneous equation.
So use \( C \) and \( D \) as the arbitrary constants.
The general solution for \( y(t) \) is:
\[
y(t) = \boxed{ \ \ }
\]
**Explanation:**
In these types of problems, we first solve the homogeneous differential equation:
\[
\frac{d^2y}{dt^2} + 13\frac{dy}{dt} + 36y = 0
\]
This will give us the complementary solution involving \( y_1(t) \) and \( y_2(t) \).
Then, we find a particular solution \( y_p(t) \) for the non-homogeneous part \( 4t^2 - 6 + 10e^{-2t} \).
Combining the complementary solution and the particular solution gives us the general solution.
**NOTE:**
In practice, solving each part requires additional steps, such as finding the roots of the characteristic equation for the complementary part and applying methods like undetermined coefficients or variation of parameters for the particular solution. This box is reserved for inputting the final general solution after completing those steps.
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