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 in order to achieve the valid solution of the differential equations, such as product rule, quotient rule and chain rule particularly.
Many students missed applying the chain rule which resulted in unexpected outcome of the differential equation.
Take a look at the following example for a typical differential equation.

Worked Examples of Differential Equations

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 \(x\).

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.

(b)   A solution of the differential equation is given by \(y = e^{kt}\) for some values \(k\). Find the possible values of \(k\).

(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 \(B\).

You can find more details of how to perform implicit differentiation.

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