Question

Given that y = 1 when x = 0, solve the differential equation dy/dx = y(4-y), obtaining an expression for y in terms of x.

Answer 1

dy/dx = y(4-y)
dy /( y (4-y) ) = dx

Answer 2

$$
\Large \int \frac{dy}{y (4-y)} = \int dx
$$

Answer 3

$$\Large \int \frac{dy}{y (4-y)} = \Large \int \frac{1}{4y} - \frac{1}{4(y-4)}~dy\\

$$

Answer 4

Uh huh.

Answer 5

Yup. Then?

Answer 6

$$
\Large \int \frac{dy}{y (4-y)}= \int dx\\
\therefore
\Large \int \frac{1}{4y} - \frac{1}{4(y-4)}~dy=\int dx \\
\therefore
\Large \frac{1}{4}~ln|y| - \frac{1}{4}~ ln |y-4| = x+C\\
\therefore \Large \frac{1}{4}~ln|\frac{y}{y-4}| = x+C\\
\therefore \Large ln|\frac{y}{y-4}| = 4\cdot x+4C\\
\therefore \Large ln|\frac{y}{y-4}| = 4x+C^{*}\\[0.1 in]
\therefore \LARGE e^{ \Large ln|\frac{y}{y-4}|} = e^{4x+C^{*}}\\
\therefore \Large\frac{y}{y-4} = e^{4x+C^{*}}\

$$

Answer 7

C * is the same as a constant

Answer 8

so we can just call it C

Answer 9

Okay.

Answer 10

$$
\Large \int \frac{dy}{y (4-y)}= \int dx\\
\therefore
\Large \int \frac{1}{4y} - \frac{1}{4(y-4)}~dy=\int dx \\
\therefore
\Large \frac{1}{4}~ln|y| - \frac{1}{4}~ ln |y-4| = x+C\\
\therefore \Large \frac{1}{4}~ln|\frac{y}{y-4}| = x+C\\
\therefore \Large ln|\frac{y}{y-4}| = 4\cdot x+4C\\
\therefore \Large ln|\frac{y}{y-4}| = 4x+C^{*}\\[0.1 in]
\therefore \LARGE e^{ \Large ln|\frac{y}{y-4}|} = e^{4x+C^{*}}\\
\therefore \Large\frac{y}{y-4} = e^{4x+C}\\
\rm flip both sides \\
\therefore \Large\frac{y-4}{y} = \frac{1}{e^{4x+C}}\\
\therefore \Large \frac{y}{y} - \frac{4}{y} = \frac{1}{e^{4x+C}}\\

\therefore \Large 1 - \frac{4}{y} = \frac{1}{e^{4x+C}}\\

$$

Answer 11

Okay.

Answer 12

Do we put in the values alrd

Answer 13

Nice.

Answer 14

How do Im arrange?

Answer 15

Thanks. I got it alrd. Could help men itch differentiation of y = x + cos2x?

Answer 16

Is it1 - sin 2x?

Answer 17

$$


\Large \int \frac{dy}{y (4-y)}= \int dx\\
\therefore
\Large \int \frac{1}{4y} - \frac{1}{4(y-4)}~dy=\int dx \\
\therefore
\Large \frac{1}{4}~ln|y| - \frac{1}{4}~ ln |y-4| = x+C\\
\therefore \Large \frac{1}{4}~ln|\frac{y}{y-4}| = x+C\\
\therefore \Large ln|\frac{y}{y-4}| = 4\cdot x+4C\\
\therefore \Large ln|\frac{y}{y-4}| = 4x+C^{*}\\[0.1 in]
\therefore \LARGE e^{ \Large ln|\frac{y}{y-4}|} = e^{4x+C}\\
\therefore \Large|\frac{y}{y-4}| = e^{4x+C}\\
\therefore \Large |\frac{y}{y-4}| = e^{4x}\cdot e^C\\
\large \rm since~ e^{constant}= constant \\[0.1 in]
\therefore \Large |\frac{y}{y-4}| = C ~e^{4x}\\
\text{ the sign of the absolute value} \pm \text{gets absorbed by the constant C } \\
\therefore \Large \frac{y}{y-4} = C ~e^{4x}\\

$$

Answer 18

you need to use chain rule
y = x + cos2x
dy/dx = 1 - sin(2x) * 2