Solve the initial value problem for x as a function of t.
-$$( t + 4 ) \frac { d x } { d t } = x ^ { 2 } + 1 , \quad t - 4 , x ( 1 ) = \tan 1$$
A) $x = \tan [ \ln | t + 4 | - \ln 5 ]$
B) $x = \tan [ \ln | t + 4 | - \ln 5 + 1 ]$
C) $x = \tan ^ { - 1 } [ \ln | t + 4 | - \ln 5 + 1 ]$
D) $x = \frac { 1 } { t + 4 } - \frac { 1 } { t - 5 }$

Question

Solve the initial value problem for x as a function of t.
-$$( t + 4 ) \frac { d x } { d t } = x ^ { 2 } + 1 , \quad t > - 4 , x ( 1 ) = \tan 1$$
A) $x = \tan [ \ln | t + 4 | - \ln 5 ]$
B) $x = \tan [ \ln | t + 4 | - \ln 5 + 1 ]$
C) $x = \tan ^ { - 1 } [ \ln | t + 4 | - \ln 5 + 1 ]$
D) $x = \frac { 1 } { t + 4 } - \frac { 1 } { t - 5 }$

Quizplus · Accepted Answer

The Answer of Solve the initial value problem for x as a function of t.
-$$( t + 4 ) \frac { d x } { d t } = x ^ { 2 } + 1 , \quad t > - 4 , x ( 1 ) = \tan 1$$
A) $x = \tan [ \ln | t + 4 | - \ln 5 ]$
B) $x = \tan [ \ln | t + 4 | - \ln 5 + 1 ]$
C) $x = \tan ^ { - 1 } [ \ln | t + 4 | - \ln 5 + 1 ]$
D) $x = \frac { 1 } { t + 4 } - \frac { 1 } { t - 5 }$

Solve the Initial Value Problem for X as a Function (t+4)dxdt=x2+1,t>−4,x(1)=tan⁡1( t + 4 ) \frac { d x } { d t } = x ^ { 2 } + 1 , \quad t > - 4 , x ( 1 ) = \tan 1(t+4)dtdx​=x2+1,t>−4,x(1)=tan1

Solve the Initial Value Problem for X as a Function $( t + 4 ) \frac { d x } { d t } = x ^ { 2 } + 1 , \quad t > - 4 , x ( 1 ) = \tan 1$