Answer:
[tex]\displaystyle \large{y = Ce^{4x}}[/tex] where C is an arbitrary constant
Step-by-step explanation:
We are given the first-order ordinary differential equation:
[tex]\displaystyle \large{\frac{dy}{dx}=4y}[/tex]
(i) y = 0 is a solution since when y = 0, dy/dx = 0.
(ii) For y doesn’t equal 0, see below.
We can not integrate both sides with respect to x directly since the RHS (Right-Handed Side) is with y-term. Therefore, we’ll have to use the separable method to separate y-term with dy and x-term with dx.
First, move dx to multiply with 4y.
[tex]\displaystyle \large{dy=4ydx}[/tex]
Now that we have this, we will move y-term to divide dy so we’ll have in the form of f(y)dy and g(x)dx.
[tex]\displaystyle \large{\frac{1}{y}dy = 4dx}[/tex]
Now, we are able to integrate both sides with respect to their own terms. The LHS (Left-Handed Side) is integrated with respect to y while the RHS is integrated with respect to x.
[tex]\displaystyle \large{\int \frac{1}{y}dy = \int 4dx}\\\displaystyle \large{\ln |y|= 4x + C}[/tex]
Where C is an arbitrary constant, technically that’s a final answer (general solution) but I’ll simplify in term of y in case if you need it.
Now to simplify the equation in term of y, you must know or recall how to convert logarithm to exponential.
Thus:
[tex]\displaystyle \large{e^{4x+C_1}=|y|}\\\displaystyle \large{\pm e^{4x+C_1}=y}\\\displaystyle \large{y= \pm e^{C_1}} \cdot e^{4x}}[/tex]
Let [tex]\displaystyle \large{\pm e^{C_1}}[/tex] be C then we have [tex]\displaystyle \large{y = Ce^{4x}}[/tex] (C is an arbitrary constant other than 0)
And if C = 0 then we get y = 0 which satisfies the first condition (i).
Hence, the general solution is [tex]\displaystyle \large{y=Ce^{4x}}[/tex] where C is an arbitrary constant.
