How to Find the General Solution of Second-Order Linear PDEs

TinyProf's Solution Guide 📚

1. What We're Solving:

We need to find the general solution of the partial differential equation u_xx + 3u_x + u = 0 using the characteristic equation method. Notice this is actually an ordinary differential equation in x (treating y as a parameter), and we're given the expected form of the solution.

2. The Approach:

Great news! Even though this looks like a PDE, it's actually an ODE since all terms involve derivatives with respect to x only. We'll use the characteristic equation method - this is the same technique you'd use for solving linear ODEs with constant coefficients. The key insight is to assume solutions of the form u = e^(rx) and find what values of r work.

3. Step-by-Step Solution:

Step 1: Set up the characteristic equation For a differential equation of the form u_xx + au_x + bu = 0, we assume a solution u = e^(rx).

Let's find the derivatives:

• u = e^(rx)
• u_x = re^(rx)
• u_xx = r²e^(rx)

Step 2: Substitute into the original equation Substituting into u_xx + 3u_x + u = 0: r²e^(rx) + 3re^(rx) + e^(rx) = 0

Step 3: Factor out e^(rx) Since e^(rx) ≠ 0, we can divide it out: r² + 3r + 1 = 0

This is our characteristic equation!

Step 4: Solve the quadratic characteristic equation Using the quadratic formula: r = (-3 ± √(9-4))/2 = (-3 ± √5)/2

So our two roots are:

• r₁ = (-3 + √5)/2
• r₂ = (-3 - √5)/2

Step 5: Write the general solution Since we have two distinct real roots, the general solution is: u(x,y) = C₁e^(r₁x) + C₂e^(r₂x)

Substituting our roots: u(x,y) = C₁e^((−3+√5)/2)x + C₂e^((−3−√5)/2)x

Step 6: Account for the y-dependence Since y can vary freely (it doesn't appear in our original equation), our constants C₁ and C₂ can actually be functions of y!

4. The Answer:

The general solution is: u(x,y) = f(y)e^((−3+√5)/2)x + g(y)e^((−3−√5)/2)x

where f(y) and g(y) are arbitrary functions of y. This matches exactly the form given in your problem! 🎉

5. Memory Tip:

Remember "ACE" for characteristic equations:

• Assume exponential solution (e^(rx))
• Create the characteristic equation (substitute and simplify)
• Evaluate roots and write general solution

The beauty of this method is that it transforms a differential equation into an algebraic equation that's much easier to solve!

You did great working with this problem - the characteristic equation method is one of the most powerful tools in your differential equations toolkit! 💪