Given the differential equation

$5t^3 \frac{dx}{dt} +2x-2=0$

The solution is as follows:

$5t^3 \frac{dx}{dt} +2x-2=0 \\ \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\ \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\ \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\ \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\ \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\ \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\ \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\ \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1$

3 0

3 years ago

Find the area and perimeter of the figure below, explain and show work pls

Stels [109]

This is my work and only part 1

7 0

3 years ago

Read 2 more answers

Please Help! Thanks to whoever can answer this!

Illusion [34]

Here, we are missing the slope and the y intercept.

Lets look for the y intercept first.

Where does it cross the y axis? At 3!!

3 is the y intercept.

Now, find the slope by using the slope formula.

$m = \frac{y2 - y1}{x2 - x1}$
(0,3) and (2,1)

m = 1-3/2-0 = -2/2 = -2/2 = -1
m = -1

Okay, so we found our slope, now just write the equation.

Answer: y = -1x + 3

8 0

3 years ago