Question

Let \(y(x)\) be the solution of the differential equation \(y'' - 4y' - 12y = 3e^{5x}\) satisfying \(y(0)=\frac{18{7}\) and \(y'(0)=-\frac{1}{7}\). Then \(y(1)\) is ______________ (rounded off to nearest integer).}

Hint:

Always check the particular solution guess \(Ae^{5x}\) when RHS is exponential.

Answer 1

Correct Answer: 339

The homogeneous equation is: \[ y_h = C_1 e^{6x} + C_2 e^{-2x} \] A particular solution is assumed as: \[ y_p = A e^{5x} \] Substituting into the DE: \[ (25A - 20A - 12A)e^{5x} = 3e^{5x} \] \[ -7A = 3 \Rightarrow A = -\frac{3}{7} \] Thus the general solution is: \[ y = C_1 e^{6x} + C_2 e^{-2x} - \frac{3}{7}e^{5x} \] Apply initial conditions: At \(x=0\): \[ C_1 + C_2 - \frac{3}{7} = \frac{18}{7} \] \[ C_1 + C_2 = 3 \] Differentiate: \[ y' = 6C_1 e^{6x} - 2C_2 e^{-2x} - \frac{15}{7}e^{5x} \] At \(x=0\): \[ 6C_1 - 2C_2 - \frac{15}{7} = -\frac{1}{7} \] \[ 6C_1 - 2C_2 = 2 \] Solving, \[ C_1 = 1,\qquad C_2 = 2 \] Now evaluate at \(x=1\): \[ y(1) = e^6 + 2e^{-2} - \frac{3}{7}e^5 \approx 340 \] Thus, the answer lies in the range: \[ \boxed{339\ \text{to}\ 341} \]
Final Answer: 339–341