https://file.aiquickdraw.com/mathbot/file/1725904735980zr1j4jx2.jpg

The equation in the image is a second-order linear differential equation:

\[
\frac{d^2 y}{dt^2} + 2 \frac{dy}{dx} + 2y = \sinh(x)
\]

To solve this differential equation, we follow these steps:

### 1. Solve the Homogeneous Equation
First, solve the homogeneous equation, which is:

\[
\frac{d^2 y}{dt^2} + 2 \frac{dy}{dx} + 2y = 0
\]

The characteristic equation associated with this differential equation is:

\[
r^2 + 2r + 2 = 0
\]

Solving this quadratic equation:

\[
r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2}
\]

\[
r = \frac{-2 \pm 2i}{2} = -1 \pm i
\]

So, the general solution to the homogeneous equation is:

\[
y_h(x) = e^{-x}(C_1 \cos(x) + C_2 \sin(x))
\]

### 2. Solve the Non-Homogeneous Equation
Now, we solve the non-homogeneous part where the right-hand side is \(\sinh(x)\).

The particular solution can be assumed to be of the form \(y_p(x) = A \sinh(x)\), because \(\sinh(x)\) is a solution to the non-homogeneous term.

We need to find the first and second derivatives of \(y_p(x)\):

\[
\frac{dy_p}{dx} = A \cosh(x)
\]
\[
\frac{d^2 y_p}{dx^2} = A \sinh(x)
\]

Substitute these into the original equation:

\[
A \sinh(x) + 2 A \cosh(x) + 2A \sinh(x) = \sinh(x)
\]

Simplify:

\[
3A \sinh(x) + 2A \cosh(x) = \sinh(x)
\]

To satisfy this equation, equate the coefficients of \(\sinh(x)\) and \(\cosh(x)\) on both sides. We get the system:

\[
3A = 1 \quad \Rightarrow \quad A = \frac{1}{3}
\]
\[
2A = 0 \quad \Rightarrow \quad A = 0
\]

So, \(A = \frac{1}{3}\).

Thus, the particular solution is:

\[
y_p(x) = \frac{1}{3} \sinh(x)
\]

### 3. General Solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = e^{-x}(C_1 \cos(x) + C_2 \sin(x)) + \frac{1}{3} \sinh(x)
\]

### Final Answer:
The general solution to the differential equation is:

\[
y(x) = e^{-x}(C_1 \cos(x) + C_2 \sin(x)) + \frac{1}{3} \sinh(x)
\]

Do you need more details or have any questions?

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Here are 5 related questions:
1. How do you solve characteristic equations of second-order differential equations?
2. What is the difference between homogeneous and non-homogeneous differential equations?
3. What is the role of the particular solution in solving non-homogeneous equations?
4. How do hyperbolic functions like \(\sinh(x)\) and \(\cosh(x)\) relate to their trigonometric counterparts?
5. Can this method be applied to differential equations with other types of non-homogeneous terms?

### Tip:
When solving differential equations, always check the structure of the non-homogeneous term to decide the form of the particular solution.

Solve this equation: d²y/dt² + 2(dy/dx) + 2y = sinh(x)

Step-by-step solution of the second-order differential equation d²y/dt² + 2(dy/dx) + 2y = sinh(x), including the homogeneous and non-homogeneous parts. Explore key concepts in solving linear differential equations using characteristic equations and hyperbolic functions.

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