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