Given a function $ y(x) $ satisfying the differential equation $$ y'' - 0.25y = 0, $$ with initial conditions $ y(0) = 1 $; $ y'(0) = 1 $, what is the value of $ y(\log_e 100) $? here, y' and y'' are the first and second derivatives of y, respectively.

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Given a function $ y(x) $ satisfying the differential equation $$ y'' - 0.25y = 0, $$ with initial conditions $ y(0) = 1 $; $ y'(0) = 1 $, what is the value of $ y(\log_e 100) $?  here, y' and y'' are the first and second derivatives of y, respectively.

cdquestions Admin · Accepted Answer

We are given the second-order linear homogeneous differential equation: $$ y'' - 0.25y = 0 $$ Step 1: Solving the differential equation. The characteristic equation for the differential equation is: $$ r^2 - 0.25 = 0 $$ Solving for $ r $: $$ r = \pm \frac{1}{2} $$ Thus, the general solution to the differential equation is: $$ y(x) = A e^{x/2} + B e^{-x/2} $$ Step 2: Using the initial conditions. From the initial condition $ y(0) = 1 $: $$ A e^{0} + B e^{0} = 1 \quad \Rightarrow \quad A + B = 1 \quad {(1)} $$ From the initial condition $ y'(0) = 1 $, we first compute the derivative: $$ y'(x) = \frac{A}{2} e^{x/2} - \frac{B}{2} e^{-x/2} $$ At $ x = 0 $: $$ \frac{A}{2} - \frac{B}{2} = 1 \quad \Rightarrow \quad A - B = 2 \quad {(2)} $$ Now, solving the system of equations (1) and (2): $$ A + B = 1 $$ $$ A - B = 2 $$ Adding these two equations: $$ 2A = 3 \quad \Rightarrow \quad A = \frac{3}{2} $$ Substitute $ A = \frac{3}{2} $ into equation (1): $$ \frac{3}{2} + B = 1 \quad \Rightarrow \quad B = -\frac{1}{2} $$ Step 3: Finding $ y(\log_e 100) $. Now that we have the values of $ A $ and $ B $, the solution to the differential equation is: $$ y(x) = \frac{3}{2} e^{x/2} - \frac{1}{2} e^{-x/2} $$ Substitute $ x = \log_e 100 $: $$ y(\log_e 100) = \frac{3}{2} e^{\log_e 100 / 2} - \frac{1}{2} e^{-\log_e 100 / 2} $$ Since $ \log_e 100 = 2 \log_e 10 $, we have: $$ y(\log_e 100) = \frac{3}{2} \times 10 - \frac{1}{2} \times \frac{1}{10} $$ Simplifying: $$ y(\log_e 100) = 15 - 0.05 = 14.95 $$ Step 4: Conclusion. The value of $ y(\log_e 100) $ is 14.95, so the correct answer is (C).

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Given a function \( y(x) \) satisfying the differential equation \[ y'' - 0.25y = 0, \] with initial conditions \( y(0) = 1 \); \( y'(0) = 1 \), what is the value of \( y(\log_e 100) \)?
here, y' and y'' are the first and second derivatives of y, respectively.

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The Correct Option is C

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Given a function \( y(x) \) satisfying the differential equation \[ y'' - 0.25y = 0, \] with initial conditions \( y(0) = 1 \); \( y'(0) = 1 \), what is the value of \( y(\log_e 100) \)? here, y' and y'' are the first and second derivatives of y, respectively.