4. The decay of a radioactive compound is described by the second order differential equation: $$ y^{\prime \prime}+y^{\prime}-20 y=0 $$ a) Find the general solution to this differential equation by solving the characteristic equation using the substitutions $$ \lambda^{2}=y^{\prime \prime}, \lambda^{1}=y^{\prime} $$ and $$ \lambda^{0}=y $$. (2 marks) b) Find the particular solution given that $$ y(0)=2 $$ and $$ y^{\prime}(0)=-10 $$.

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We are given the differential equation
$$
y^{\prime\prime} + y^{\prime} - 20y = 0.
$$

**(a) General Solution**

1. Assume a solution of the form $$ y = e^{\lambda t} $$. Then, we have:
   $$
   y^{\prime} = \lambda e^{\lambda t} \quad \text{and} \quad y^{\prime\prime} = \lambda^2 e^{\lambda t}.
   $$
2. Substituting into the differential equation gives:
   $$
   \lambda^2 e^{\lambda t} + \lambda e^{\lambda t} - 20e^{\lambda t} = 0.
   $$
3. Factor out the common term $$ e^{\lambda t} $$ (which is never zero):
   $$
   e^{\lambda t} \left( \lambda^2 + \lambda - 20 \right) = 0.
   $$
   Therefore, the characteristic equation is:
   $$
   \lambda^2 + \lambda - 20 = 0.
   $$
4. Solve the quadratic equation using the quadratic formula:
   $$
   \lambda = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-20)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 80}}{2} = \frac{-1 \pm \sqrt{81}}{2}.
   $$
5. Since $$\sqrt{81} = 9$$, the roots are:
   $$
   \lambda = \frac{-1 + 9}{2} = 4 \quad \text{and} \quad \lambda = \frac{-1 - 9}{2} = -5.
   $$
6. Hence, the general solution is:
   $$
   y(t)= C_1 e^{4t} + C_2 e^{-5t},
   $$
   where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants.

**(b) Particular Solution**

We are given the initial conditions:
$$
y(0)=2 \quad \text{and} \quad y^{\prime}(0)=-10.
$$

1. Substitute $$ t=0 $$ into the general solution:
   $$
   y(0)= C_1 e^{0}+ C_2 e^{0}= C_1+C_2=2.
   $$
   This gives the equation:
   $$
   C_1+C_2=2. \quad \text{(1)}
   $$
2. Differentiate the general solution:
   $$
   y^{\prime}(t)= 4C_1 e^{4t} -5C_2 e^{-5t}.
   $$
3. Substitute $$ t=0 $$ into this derivative:
   $$
   y^{\prime}(0)= 4C_1 - 5C_2= -10.
   $$
   This gives:
   $$
   4C_1-5C_2= -10. \quad \text{(2)}
   $$
4. Solve the system of equations (1) and (2):

From (1): 
   $$
   C_1 = 2 - C_2.
   $$
   Substitute into (2):
   $$
   4(2 - C_2) - 5C_2 = -10.
   $$
5. Simplify:
   $$
   8 - 4C_2 - 5C_2 = -10 \quad \Rightarrow \quad 8 - 9C_2 = -10.
   $$
6. Solve for $$ C_2 $$:
   $$
   -9C_2 = -18 \quad \Rightarrow \quad C_2 = 2.
   $$
7. Substitute $$ C_2 = 2 $$ into (1):
   $$
   C_1+2=2 \quad \Rightarrow \quad C_1=0.
   $$

Thus, the particular solution is:
$$
y(t)= 0\cdot e^{4t} + 2e^{-5t} = 2e^{-5t}.
$$
**(a) General Solution:** $$ y(t) = C_1 e^{4t} + C_2 e^{-5t} $$ **(b) Particular Solution:** $$ y(t) = 2e^{-5t} $$

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