Solved

Solve the Initial Value Problem for X as a Function (2t32t2+t1)dxdt=3,x(2)=0\left( 2 t ^ { 3 } - 2 t ^ { 2 } + t - 1 \right) \frac { d x } { d t } = 3 , x ( 2 ) = 0

Question 117

Multiple Choice

Solve the initial value problem for x as a function of t.
- (2t32t2+t1) dxdt=3,x(2) =0\left( 2 t ^ { 3 } - 2 t ^ { 2 } + t - 1 \right) \frac { d x } { d t } = 3 , x ( 2 ) = 0


A) x=lnt112lnt2+12+12ln4.5x = \ln | t - 1 | - \frac { 1 } { 2 } \ln \left| t ^ { 2 } + \frac { 1 } { 2 } \right| + \frac { 1 } { 2 } \ln 4.5
B) x=lnt12tan12t+2tan122x = \ln | t - 1 | - \sqrt { 2 } \tan ^ { - 1 } \sqrt { 2 } t + \sqrt { 2 } \tan ^ { - 1 } 2 \sqrt { 2 }
C) x=lnt1tan12tlnt2+12+tan122+ln4.5x = \ln | t - 1 | - \tan ^ { - 1 } \sqrt { 2 } t - \ln \left| t ^ { 2 } + \frac { 1 } { 2 } \right| + \tan ^ { - 1 } 2 \sqrt { 2 } + \ln 4.5
D) x=lnt12tan12t12lnt2+12+2tan122+12ln4.5x = \ln | t - 1 | - \sqrt { 2 } \tan ^ { - 1 } \sqrt { 2 } t - \frac { 1 } { 2 } \ln \left| t ^ { 2 } + \frac { 1 } { 2 } \right| + \sqrt { 2 } \tan ^ { - 1 } 2 \sqrt { 2 } + \frac { 1 } { 2 } \ln 4.5

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions