Question

Differential equations by separation of variables

dy/dx= 4y
(help with how to solve the differential equations) ...?

Answer 1

`\displaystyle \large{y = Ce^(4x)}` where C is an arbitrary constant

Explanation:

We are given the first-order ordinary differential equation:

`\displaystyle \large{(dy)/(dx)=4y}`

(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.

`\displaystyle \large{dy=4ydx}`

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.

`\displaystyle \large{(1)/(y)dy = 4dx}`

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.

`\displaystyle \large{\int (1)/(y)dy = \int 4dx}\\\displaystyle \large\ln`

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:

`\displaystyle \largey\\\displaystyle \large{\pm e^(4x+C_1)=y}\\\displaystyle \large{y= \pm e^(C_1)} \cdot e^(4x)}`

Let
`\displaystyle \large{\pm e^(C_1)}` be C then we have
`\displaystyle \large{y = Ce^(4x)}` (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
`\displaystyle \large{y=Ce^(4x)}` where C is an arbitrary constant.