Solve the differential equation with the given initial condition.
-$y ^ { \prime } = y ( t - 2 ) ; y ( 0 ) = 1$
A) y = $e \left[ \left( t ^ { 2 } / 2 \right) - 2 t \right]$ - 2t + 1
B) y = $\frac { t ^ { 2 } } { 2 }$ - 2t
C) y = $e \left[ \left( t ^ { 2 } / 2 \right) - 2 t \right]$
D) y = 0
E) none of these

Question

Solve the differential equation with the given initial condition. 
-$y ^ { \prime } = y ( t - 2 ) ; y ( 0 ) = 1$
A) y = $e \left[ \left( t ^ { 2 } / 2 \right) - 2 t \right]$ - 2t + 1
B) y = $\frac { t ^ { 2 } } { 2 }$ - 2t
C) y = $e \left[ \left( t ^ { 2 } / 2 \right) - 2 t \right]$
D) y = 0
E) none of these

Quizplus · Accepted Answer

The differential equation is separable. Solving it gives $ y = Ce^{(t^2/2) - 2t} $. Using the initial condition $ y(0) = 1 $, we find $ C = 1 $. Thus, the solution is $ y = e^{(t^2/2) - 2t} $.

Solve the Differential Equation with the Given Initial Condition y′=y(t−2);y(0)=1y ^ { \prime } = y ( t - 2 ) ; y ( 0 ) = 1y′=y(t−2);y(0)=1

Solve the Differential Equation with the Given Initial Condition $y ^ { \prime } = y ( t - 2 ) ; y ( 0 ) = 1$