The partial differential equation (PDE) given is:

\[
u_{yy} + 9u_x + 7u_y = \cos(y)
\]

This is a second-order linear PDE with mixed derivatives. To solve it, let's use the method of characteristics, or, if suitable, we can try an ansatz (a guessed form) for the solution.

### Step 1: Analyze the PDE

The given PDE is not homogeneous due to the \(\cos(y)\) on the right-hand side. We'll look for a particular solution and then the general solution to the associated homogeneous equation.

### Step 2: Solve the Homogeneous Equation

The homogeneous equation associated with this PDE is:

\[
u_{yy} + 9u_x + 7u_y = 0
\]

To solve this, we assume a solution of the form \(u(x, y) = e^{\alpha x + \beta y}\).

Substituting into the homogeneous PDE:

\[
\alpha^2 e^{\alpha x + \beta y} + 9\alpha e^{\alpha x + \beta y} + 7\beta e^{\alpha x + \beta y} = 0
\]

This simplifies to:

\[
\alpha^2 + 9\alpha + 7\beta = 0
\]

This quadratic equation in \(\alpha\) can be solved, but let's now consider the particular solution due to the non-homogeneous term.

### Step 3: Find the Particular Solution

Since the non-homogeneous term is \(\cos(y)\), let's guess a particular solution of the form:

\[
u_p(y) = A \cos(y) + B \sin(y)
\]

Substituting \(u_p(y)\) into the original equation:

\[
u_{yy} + 9u_x + 7u_y = \cos(y)
\]

We find that:

\[
-u_p(y) + 7u_y = \cos(y)
\]

Substituting \(u_p(y)\) gives us:

\[
-(-A \cos(y) - B \sin(y)) + 7(-A \sin(y) + B \cos(y)) = \cos(y)
\]

Equating coefficients for \(\cos(y)\) and \(\sin(y)\), we get two equations:

\[
A + 7B = 1
\]
\[
-B + 7A = 0
\]

Solving these simultaneous equations, we get:

\[
B = \frac{1}{50}, \quad A = \frac{7}{50}
\]

Thus, the particular solution is:

\[
u_p(y) = \frac{7}{50} \cos(y) + \frac{1}{50} \sin(y)
\]

### Step 4: General Solution

The general solution \(u(x, y)\) is the sum of the general solution to the homogeneous equation and the particular solution:

\[
u(x, y) = u_h(x, y) + u_p(y)
\]

Given \(u_h(x, y)\) is the solution to the homogeneous PDE, the final solution is:

\[
u(x, y) = e^{\alpha x + \beta y} + \frac{7}{50} \cos(y) + \frac{1}{50} \sin(y)
\]

Where \(\alpha\) and \(\beta\) satisfy the quadratic equation \(\alpha^2 + 9\alpha + 7\beta = 0\).

### Conclusion

This general form can be further simplified depending on initial or boundary conditions provided.

Would you like more details on solving the quadratic equation for \(\alpha\) and \(\beta\), or how this solution could be applied in specific contexts?

### Related Questions
1. How do you solve non-homogeneous linear PDEs?
2. What is the method of characteristics for solving PDEs?
3. How can boundary conditions affect the solution of a PDE?
4. What are the applications of PDEs in physics and engineering?
5. How do Fourier series help in solving PDEs?
6. What is the significance of homogeneous solutions in differential equations?
7. Can all PDEs be solved analytically?
8. What are common methods for finding particular solutions to non-homogeneous PDEs?

### Tip
When dealing with non-homogeneous PDEs, always remember to find the particular solution first before addressing the homogeneous part of the equation.

Learn how to solve the partial differential equation (PDE) uyy + 9ux + 7uy = cos(y) using the method of characteristics and finding particular solutions. This problem involves second-order PDEs and requires an understanding of non-homogeneous differential equations. Suitable for advanced college-level mathematics.

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