# Finding a Function from Differential Equation

The solution of a differential equation is to find an expression without $\displaystyle \frac{d}{dx}$ notations using given conditions.
Note that the proper rules must be in place to achieve a valid solution of the differential equations, such as the product, quotient, and chain rules.
Many students missed applying the chain rule, resulting in an unexpected differential equation outcome.
Take a look at the following example for a typical differential equation.

## Question 1

A function $f(x)$ is defined for all real $x$ and $y$ such that $\require{color} f(x + y) = f(x)f(y) \text{ , and } f'(0) = 2$.
Find $f(x)$ in terms of $x$.

\begin{aligned} \require{AMSsymbols} \require{color} f(y) &= f(0)f(y) &\color{green} \text{substitute } x=0 \\ \therefore f(0)&=1 &\color{green} (1) \\ \frac{d}{dx}f(x + y) \times \frac{d}{dx}(x + y) &= \frac{d}{dx}f(x) \times f(y) + f(x) \times \frac{d}{dx}f(y) &\color{green} \text{differentiate both sides} \\ \frac{d}{dx}f(x + y) \times (1 + \frac{dx}{dy}) &= \frac{d}{dx}f(x) \times f(y) + f(x) \times \frac{d}{dx}f(y) \times \frac{dy}{dx} &\color{green} \frac{dy}{dx} = y’ \\ f'(x + y) \times (1 + y’) &= f'(x) \times f(y) + f(x) \times f'(y) \times y’ \\ f'(y) \times (1 + y’) &= f'(0) \times f(y) + f(0) \times f'(y) \times y’ &\color{green} \text{substitute } x=0 \\ f'(y) \times (1 + y’) &= 2 \times f(y) + f(0) \times f'(y) \times y’ &\color{green} \text{substitute } f'(0) = 2 \\ f'(y) + f'(y) y’ &= 2 f(y) + 1 \times f'(y) y’ &\color{green} \text{substitute } f(0) = 1 \\ f'(y) &= 2 f(y) \\ \frac{f'(y)}{f(y)} &=2 \\ \int \frac{f'(y)}{f(y)} \,dy &= \int 2 \,dy &\color{green} \text{integrate both sides} \\ \log_e f(y) &= 2y + C &\color{green} (2) \\ \log_e f(0) &= 2 \times 0 + C &\color{green} \text{substitute } y = 0 \\ \log_e 1 &= C &\color{green} \text{substitute } f(0) = 1 \\ \log_e f(y) &= 2y &\color{green} (2) \\ f(y) &= e^{2y} \\ \therefore f(x) &= e^{2x} \\ \end{aligned}

## Question 2

Consider a differential equation, $\displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$.

(a)   Show that, if $y = f(t)$ and $y = g(t)$ are both solutions to the differential equation and $A$ and $B$ are constants, then $y = Af(t) + Bg(t)$ is a solution.

If $y = f(t)$ is a solution of $\require{AMSsymbols} \require{color} \displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$, then $\displaystyle \require{AMSsymbols} \require{color} \frac{d^2}{dt}f(t) +3\frac{d}{dt}f(t) + 2f(t) = 0 \color{green} \cdots (1)$
If $y = g(t)$ is a solution of $\require{color} \displaystyle \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$, then $\displaystyle \require{AMSsymbols} \require{color} \frac{d^2}{dt}g(t) +3\frac{d}{dt}g(t) + 2g(t) = 0 \color{green} \cdots (2)$
\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} y &= Af(t) + Bg(t) \cdots \color{green} (3) \\ \frac{dy}{dt} &= A\frac{d}{dt}f(t)+B\frac{d}{dt}g(t) \cdots \color{green} (4) \\ \frac{d^2y}{dt^2} &= A\frac{d^2}{dt^2}f(t)+B\frac{d^2}{dt^2}g(t) \cdots \color{green} (5) \end{aligned}
\begin{aligned} \displaystyle \require{AMSsymbols}\require{color} A\frac{d^2}{dt^2}f(t)+ B\frac{d^2}{dt^2}g(t) + 3A\frac{d}{dt}f(t)+3B\frac{d}{dt}g(t) + 2Af(t) + 2Bg(t) &= 0 &\color{green} (3), (4), (5) \text{ into } \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0 \\ A \Big[\frac{d^2}{dt^2}f(t)+3\frac{d}{dt}f(t)+ 2f(t)\Big]+ B \Big[\frac{d^2}{dt^2}g(t) +3\frac{d}{dt}g(t) + 2g(t)\Big] &= 0 \\ A \times 0 + B \times 0 &= 0 &\color{green} \text{by (1) and (2)} \\ \therefore y = Af(t) + Bg(t) \text{ is a solution of } \frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y &= 0 \end{aligned}

(b)   This gives a solution of the differential equation $y = e^{kt}$ for some values $k$. Find the possible values of $k$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} k^2e^{kt} + 3ke^{kt} + 2e^{kt} &= 0 \\ e^{kt}(k^2 + 3k + 2) &= 0 \\ e^{kt}(k + 1)(k + 2) &= 0 \\ \therefore k &= -1 \text{ or } -2 &\color{green} e^{kt} \ne 0 \\ \end{aligned}

(c)   A solution of the differential equation is $y = Ae^{-2t} + Be^{-t}$. When $t=0$, it is given that $y=0$ and $\displaystyle \frac{dy}{dx} = 1$. Find the values of $A$ and $B$.

\begin{aligned} \displaystyle\require{AMSsymbols} \require{color} 0 &= Ae^{-2 \times 0} + Be^{-0} &\color{green} t=0 \text{ and } y=0 \\ 0 &= A + B &\color{green} (1) \\ \frac{dy}{dx} &= -2Ae^{-2t}-Be^{-t} \\ 1 &= -2Ae^{-2 \times 0}-Be^{-0} &\color{green} t=0 \text{ and } \frac{dy}{dx} = 1 \\ 1 &= -2A-B &\color{green} (2) \\ \therefore A &= -1 \text{ and } B = 1 &\color{green} \text{solve (1) and (2) simultaneously} \end{aligned}