Discussion
Consider a second order linear ordinary differential equation \(\frac{d^{2}y}{dx^{2}} - 4\frac{dy}{dx} + 4y = 0\), with the boundary conditions \(y(0) = 1;\ \left.\frac{dy}{dx}\right|_{x=0} = 1\). The value of y at x =1 is
Answer: A
The value of \(y\) at \(x = 1\) is 0.
The characteristic equation is \(m^{2} - 4m + 4 = 0\), that is \((m - 2)^{2} = 0\), a repeated root \(m = 2\). So the general solution is \(y = (A + Bx)e^{2x}\).
\[ \begin{aligned} y(0) &= A = 1 \\ y'(x) &= (2A + B + 2Bx)e^{2x} \\ y'(0) &= 2A + B = 1 \Rightarrow B = -1 \\ y(1) &= (1 - 1)e^{2} = 0 \end{aligned} \]
The option \(e^{2}\) is the value of \(e^{2x}\) alone, which satisfies \(y(0) = 1\) but not the slope condition, and neither 1 nor \(e\) fits both conditions.
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