1. **Problem Statement:** Solve the PDEs given in 3a and 3b. 2. **Recall the PDEs:** (a) $$\frac{\partial u}{\partial x} + u = \frac{\partial u}{\partial t}$$ with initial condition $$u = 4e^{-5x}$$ when $$t=0$$. (b) $$\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$$ with initial condition $$u(x,0) = 4e^{-2x}$$. --- ### 3a Solution: 3. Rearrange the PDE: $$\frac{\partial u}{\partial t} - \frac{\partial u}{\partial x} = -u$$ 4. Use method of characteristics or treat as first order linear PDE. 5. Consider $u$ as a function of $x$ and $t$, solve by integrating factor: Rewrite as: $$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} - u$$ 6. Fix $x$, treat as ODE in $t$: $$\frac{d u}{d t} = \frac{\partial u}{\partial x} - u$$ But since $u$ depends on both $x$ and $t$, use separation of variables or try solution form. 7. Try solution $u(x,t) = X(x)T(t)$. Substitute into PDE: $$X(x) T'(t) + X(x) T(t) = X'(x) T(t)$$ Divide both sides by $X(x) T(t)$: $$\frac{T'(t)}{T(t)} + 1 = \frac{X'(x)}{X(x)}$$ 8. Since LHS depends only on $t$ and RHS only on $x$, both equal a constant $-\lambda$: $$\frac{T'(t)}{T(t)} + 1 = -\lambda$$ $$\frac{X'(x)}{X(x)} = -\lambda$$ 9. Solve for $X(x)$: $$X'(x) = -\lambda X(x) \implies X(x) = A e^{-\lambda x}$$ 10. Solve for $T(t)$: $$T'(t) = (-\lambda -1) T(t) \implies T(t) = B e^{(-\lambda -1) t}$$ 11. General solution: $$u(x,t) = C e^{-\lambda x} e^{(-\lambda -1) t} = C e^{-\lambda x - (\lambda +1) t}$$ 12. Apply initial condition $t=0$: $$u(x,0) = 4 e^{-5x} = C e^{-\lambda x}$$ So, $$C e^{-\lambda x} = 4 e^{-5x} \implies C=4, \lambda=5$$ 13. Final solution: $$u(x,t) = 4 e^{-5x - (5+1) t} = 4 e^{-5x - 6t}$$ --- ### 3b Solution: 14. Given PDE: $$\frac{\partial u}{\partial x} = 2 \frac{\partial u}{\partial t} + u$$ Rewrite as: $$\frac{\partial u}{\partial x} - 2 \frac{\partial u}{\partial t} = u$$ 15. Try separation of variables $u(x,t) = X(x) T(t)$: $$X'(x) T(t) - 2 X(x) T'(t) = X(x) T(t)$$ Divide both sides by $X(x) T(t)$: $$\frac{X'(x)}{X(x)} - 2 \frac{T'(t)}{T(t)} = 1$$ 16. Rearrange: $$\frac{X'(x)}{X(x)} - 1 = 2 \frac{T'(t)}{T(t)}$$ LHS depends on $x$, RHS on $t$, so both equal constant $\mu$: $$\frac{X'(x)}{X(x)} - 1 = \mu$$ $$2 \frac{T'(t)}{T(t)} = \mu$$ 17. Solve for $X(x)$: $$X'(x) = (1 + \mu) X(x) \implies X(x) = A e^{(1+\mu) x}$$ 18. Solve for $T(t)$: $$\frac{T'(t)}{T(t)} = \frac{\mu}{2} \implies T(t) = B e^{\frac{\mu}{2} t}$$ 19. General solution: $$u(x,t) = A B e^{(1+\mu) x + \frac{\mu}{2} t} = C e^{(1+\mu) x + \frac{\mu}{2} t}$$ 20. Apply initial condition $t=0$: $$u(x,0) = 4 e^{-2x} = C e^{(1+\mu) x}$$ So, $$C e^{(1+\mu) x} = 4 e^{-2x} \implies 1 + \mu = -2, \quad C=4$$ Thus, $$\mu = -3$$ 21. Final solution: $$u(x,t) = 4 e^{(1 - 3) x - \frac{3}{2} t} = 4 e^{-2x - \frac{3}{2} t}$$ --- **Final answers:** $$\boxed{u(x,t) = 4 e^{-5x - 6t} \quad \text{for 3a}}$$ $$\boxed{u(x,t) = 4 e^{-2x - \frac{3}{2} t} \quad \text{for 3b}}$$