Find the general solution of the differential equation 5e y" – 2y' + y = 1+t2° NOTE: Use C, and C2 as arbitrary constants. y(t) %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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**Problem Statement:**

Find the general solution of the differential equation:

\[ y'' - 2y' + y = \frac{5e^t}{1 + t^2} \]

**Instructions:**

NOTE: Use \( C_1 \) and \( C_2 \) as arbitrary constants.

**Solution Form:**

\[ y(t) = \]

**Explanation:**

The given differential equation is a second-order linear non-homogeneous differential equation. Solving it typically involves finding the complementary solution of the associated homogeneous equation and a particular solution for the non-homogeneous part.

The homogeneous equation is:

\[ y'' - 2y' + y = 0 \]

For the non-homogeneous term, \(\frac{5e^t}{1 + t^2}\), methods like undetermined coefficients or variation of parameters might be used to find a particular solution. Finally, the general solution is given by the sum of the complementary and particular solutions.
Transcribed Image Text:**Problem Statement:** Find the general solution of the differential equation: \[ y'' - 2y' + y = \frac{5e^t}{1 + t^2} \] **Instructions:** NOTE: Use \( C_1 \) and \( C_2 \) as arbitrary constants. **Solution Form:** \[ y(t) = \] **Explanation:** The given differential equation is a second-order linear non-homogeneous differential equation. Solving it typically involves finding the complementary solution of the associated homogeneous equation and a particular solution for the non-homogeneous part. The homogeneous equation is: \[ y'' - 2y' + y = 0 \] For the non-homogeneous term, \(\frac{5e^t}{1 + t^2}\), methods like undetermined coefficients or variation of parameters might be used to find a particular solution. Finally, the general solution is given by the sum of the complementary and particular solutions.
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