If y = f(x) is solution of $\frac{d^{2}y}{d{x}^2}=0$ with the boundary conditions y = 5 at x = 0, and $\frac{dy}{dx}=2$ at x = 10, f(15) = _____

Concept:

$\frac{d^{2}y}{d{x}^2}=0\Rightarrow D^{2}y=0$

⇒Auxiliary equation is m² = 0 ⇒m = 0, 0

Then , y = (C₁ + C₂x) e${}^{0.x}$ ⇒ y = C₁ + C₂x ……(1)

Calculation:

Applying boundary conditions

At x = 0, y = 5, ⇒ 5 = C₁ + C₂ (0) ⇒ C₁ = 5

And at X = 10, $\frac{dy}{dx}=2$

Then form the equation (1), we have

y = 5 + 2x

Then, at X = 15, y = 5 + 2 (15) = 35