Question

find the solution of the given initial value problem. $$ y^{\prime}-2 y=e^{2 x}, \quad y(0)=2 $$

Short answer

Answer: The solution for the given IVP is $y(x) = e^{2x}(x + 2)$.

Step-by-step solution

Identify the given differential equation and initial condition

The given differential equation is $$ y'(x)-2y(x)=e^{2x} $$ and the initial condition is $$ y(0) = 2 $$

Calculate the integrating factor

In our case, the function associated with y(x) is -2. Let's call this function p(x): $$ p(x) = -2 $$ Now, the integrating factor (IF) is given by the exponential of the integral of p(x) tensorwith respect to x: $$ \text{IF} = e^{\int p(x) \, dx} $$ Plugging the value of p(x) and simplify: $$ \text{IF} = e^{\int -2 \, dx} = e^{-2x}\ $$

Multiply the differential equation by the integrating factor

Now, multiply the given differential equation by the computed integrating factor: $$ e^{-2x} (y'(x) - 2y(x)) = e^{-2x} e^{2x} $$ which becomes: $$ (e^{-2x}y(x))' = 1 $$

Integrate both sides of the equation

Now, we integrate both sides of the equation with respect to x: $$ \int (e^{-2x}y(x))' \, dx = \int 1 \, dx $$ This results in $$ e^{-2x}y(x) = x + C $$

Rearrange the equation to find the general solution

Now, rearrange the equation above to find y(x): $$ y(x) = e^{2x}(x + C) $$ Now we have the general solution.

Apply the initial condition to find the particular solution

Using the initial condition y(0) = 2, we can find the particular solution: $$ 2 = e^{2\cdot0}(0 + C) $$ which gives us $$ C = 2 $$

Write down the final solution

Finally, plug the value of C back into the general solution to find the particular solution: $$ y(x) = e^{2x}(x + 2) $$ That is the solution of the given initial value problem.