Solve the second‑order linear homogeneous differential equation $y^{''} - 5 y^{'} + 6 y = 0$ and determine the solution that satisfies the initial conditions $y (0) = 2$ and $y^{'} (0) = 0$ .

hard 3 marks Class 12 · Mathematics · Differential Equations

Answer

y = 6 e^{2 x} - 4 e^{3 x}

The correct answer is $y = 6 e^{2 x} - 4 e^{3 x}$ — Hence $C_{2} = 2 - 6 = - 4$

Step-by-step solution

Assume a solution of the form $y = e^{r x}$
Substitute into the ODE: $r^{2} e^{r x} - 5 r e^{r x} + 6 e^{r x} = 0$
Divide by $e^{r x}$ (never zero): $r^{2} - 5 r + 6 = 0$
Solve quadratic: $(r - 2) (r - 3) = 0$ giving $r_{1} = 2$ , $r_{2} = 3$
General solution: $y = C_{1} e^{2 x} + C_{2} e^{3 x}$
Compute derivative $y^{'} = 2 C_{1} e^{2 x} + 3 C_{2} e^{3 x}$
Apply initial condition $y (0) = 2$ : $C_{1} + C_{2} = 2$
Apply initial condition $y^{'} (0) = 0$ : $2 C_{1} + 3 C_{2} = 0$
Solve system: from first, $C_{2} = 2 - C_{1}$ ; substitute into second: $2 C_{1} + 3 (2 - C_{1}) = 0$ => $2 C_{1} + 6 - 3 C_{1} = 0$ => $- C_{1} + 6 = 0$ => $C_{1} = 6$
Hence $C_{2} = 2 - 6 = - 4$

Common mistake: Incorrectly computing characteristic roots

Concept tested: Characteristic equation for constant coefficient ODE