Determine the unique solution of:
$$y'+4y+5\int_0^x y\,dx = e^{-x},$$ given that $y(0)=0$.
[Hint: Take the derivative of both side of the given equation before you start solving.]
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We can also solve this equation by using the Laplace transform (if the hint is not only the choice). Let $Y(s)=\cal L\{(y(x)\}$. Then the Laplace transform of either side of the equation gives $$sY(s)-y(0)+5\cal \{1*y\}=\frac{1}{s+1},$$ where $1*y$ is the convolution of the functions $1$ and $y$, which is equal to $\frac{Y(s)}{s}$. Using this and the condition $y(0)=0$ in the equation we get $$\frac{s^2+5}{s}Y(s)=\frac{1}{s+1},$$ or $$Y(s)=\frac{s}{(s+1)(s^2+5)}=-\frac{1}{6}.\frac{1}{s+1}+\frac{5}{6}.\frac{1}{s^2+5}+\frac{1}{6}.\frac{s}{s^2+5}.$$ The inverse Laplace transform yields that $y(x)=-\frac{1}{6}e^{-x}+\frac{1}{6\sqrt{5}}\sin \sqrt{5}x+\frac{1}{6}\cos\sqrt{5} x$.
If we take the derivative, $\dfrac{d}{dx}(y'+4y+5\int_0^x y\,dx = e^{-x})$, we have:
$$\tag 1 y''+4y'+5y = -e^{-x}$$
We now have a second order DEQ, so we need two initial conditions and only have one.
We can use the first IC to find the second one:
$$y'(0) + 4 y(0) + 5\int_0^0 y\,dx = e^{-0} \rightarrow y'(0) = 1$$
Now, we can use undetermined coefficients to find the homogeneous solution and guessing to find the particular solution.
For the homogeneous, we have:
$$m^2 + 4m + 5 = 0 \rightarrow m_{1,2} = -2 \pm ~i$$
This gives a homogeneous solution of:
$$y_h(x) = e^{-2x} (c_1 \cos x + c_2 \sin x)$$
For the particular solution, we guess at a solution of $y_p = ae^{-x}$, substitute and solve for $a$, yielding:
$$y_p(x) = -\dfrac{1}{2}e^{-x}$$
Our solution is:
$$y(x) = y_h(x) + y_p(x) = e^{-2x} (c_1 \cos x + c_2 \sin x)-\dfrac{1}{2}e^{-x}$$
Now, substitute in the two ICs and solve for $c_1$ and $c_2$.
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