Solve the initial-value problem for x as a function of t .

Question

Solve the initial-value problem for x as a function of t .

(t²-6t+5) dx/dt =1,(t>5,x(8)=0)

Solve the initial-value problem for x  as a function of t .

(2t³-2t²+t-1) dx/dt =6, x(2)=0

I have 0 clue how to do this, it is on my calculus homework, but we have never done problems like this.

Answer 1

The problem has switched things around on you. Normally you have y=f(x). For these cases, x = f(t)

1)

(t²-6t+5)dx/dt = 1

∫dx = ∫dt/((t²-6t+5)) = ∫dt/((t-1)(t-5))

Use partial fraction decomposition

x = ∫4/(t-5)dt - ∫4/(t-1)dt

x = 4ln(t-5)-4ln(t-1)+C = 4ln((t-5)/(t-1))+C We can use parentheses for the logs here since t>5

If x(8) = 0, C = -4ln(3/7)

2)

This one follows the same pattern,

x(t)=∫6/(2t³-2t²+t-1) dt = ∫2/(t-1) dt -4∫(t+1)/(2t²+1) dt

Just integrate the two simpler integrals and then use x(2)=0 to solve for C.