Use the Laplace transform to solve the integro-differential equation y′(t) − 2∫₀ᵗ y(τ) sin(t − τ) dτ = 1 with initial condition y(0) = −1.

Step-by-step solution

1. 1

   Recognise the integral as a convolution. The integral has the exact shape $\int_0^t y(\tau)g(t-\tau)\,d\tau=(y*g)(t)$ with $g(t)=\sin t$. That is why the Laplace transform is the right tool: the convolution theorem says $\mathcal{L}\{y*g\}=Y(s)G(s)$, so a messy integral becomes an ordinary product. With $\mathcal{L}\{\sin t\}=\tfrac{1}{s^2+1}$, the integral term transforms to $\tfrac{2Y(s)}{s^2+1}$.

2. 2

   Transform the whole equation. Using $\mathcal{L}\{y'\}=sY(s)-y(0)$ and $y(0)=-1$, and $\mathcal{L}\{1\}=\tfrac1s$:

   $sY(s)+1-\frac{2Y(s)}{s^2+1}=\frac{1}{s}.$

3. 3

   Solve algebraically for $Y(s)$. Collect the $Y$ terms over a common denominator and move the constant to the right:

   $Y(s)\cdot\frac{s(s^2+1)-2}{s^2+1}=\frac{1}{s}-1=\frac{1-s}{s}.$

   The cubic factors as $s^3+s-2=(s-1)(s^2+s+2)$, and $1-s=-(s-1)$, so the $(s-1)$ cancels — a strong sign the algebra is on track:

   $Y(s)=-\frac{s^2+1}{s\,(s^2+s+2)}.$

4. 4

   Split into partial fractions. Write $\dfrac{s^2+1}{s(s^2+s+2)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2+s+2}$. Clearing denominators gives $(A+B)s^2+(A+C)s+2A=s^2+1$, so $2A=1$, $A+C=0$, $A+B=1$, i.e. $A=\tfrac12$, $B=\tfrac12$, $C=-\tfrac12$. Hence

   $Y(s)=-\frac{1}{2s}-\frac{s-1}{2(s^2+s+2)}.$

5. 5

   Complete the square to match the shift theorem. The quadratic has complex roots, so instead of factoring it we write $s^2+s+2=\left(s+\tfrac12\right)^2+\tfrac74$ and re-express the numerator around the same shift, $s-1=\left(s+\tfrac12\right)-\tfrac32$:

   $Y(s)=-\frac{1}{2s}-\frac12\cdot\frac{s+\tfrac12}{\left(s+\tfrac12\right)^2+\left(\tfrac{\sqrt7}{2}\right)^2}+\frac34\cdot\frac{1}{\left(s+\tfrac12\right)^2+\left(\tfrac{\sqrt7}{2}\right)^2}.$

6. 6

   Invert term by term. The three pieces are the transforms of a constant, a damped cosine and a damped sine (the last one needs the factor $\tfrac{1}{\omega}=\tfrac{2}{\sqrt7}$, so $\tfrac34\cdot\tfrac{2}{\sqrt7}=\tfrac{3}{2\sqrt7}$):

   $y(t)=-\frac12-\frac12e^{-t/2}\cos\!\left(\frac{\sqrt7}{2}t\right)+\frac{3}{2\sqrt7}e^{-t/2}\sin\!\left(\frac{\sqrt7}{2}t\right).$

7. 7

   Check the solution. At $t=0$ the sine term vanishes and $y(0)=-\tfrac12-\tfrac12=-1$, matching the initial condition. Substituting the closed form back and evaluating $y'(t)-2\int_0^t y(\tau)\sin(t-\tau)\,d\tau$ numerically at several values of $t$ returns $1.000000$ every time, so the answer satisfies the original equation, not just the initial condition.