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

What is the value of 34 + 3.5(−44÷2)? a.−111 b.−43 c.43 d.111

mart [117]

Answer:

Step-by-step explanation:

-44÷2=-22

34+3.5*-22= -43

5 0

2 years ago