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).}

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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

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).}

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Correct Answer: 339

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