Given the differential equation with the given initial condition: $\mathrm { y } ^ { \prime } = \mathrm { t } \cos \mathrm { t } ; \mathrm { y } ( 0 ) = 0$ is this the solution $y = t \sin t + \cos t - 1 ?$

Question

Quizplus · Accepted Answer

To verify if $y = t \sin t + \cos t - 1$ is the solution to the differential equation $y' = t \cos t$ with the initial condition $y(0) = 0$, we differentiate $y$ with respect to $t$ and check if it satisfies the given differential equation and initial condition. Differentiating $y = t \sin t + \cos t - 1$ gives $y' = \sin t + t \cos t - \sin t = t \cos t$, which matches the given differential equation. Substituting $t = 0$ into $y = t \sin t + \cos t - 1$ gives $y(0) = 0 \cdot \sin(0) + \cos(0) - 1 = 0$, which satisfies the initial condition. Therefore, the given function is indeed a solution to the differential equation with the specified initial condition.

Given the Differential Equation with the Given Initial Condition y′=tcos⁡t;y(0)=0\mathrm { y } ^ { \prime } = \mathrm { t } \cos \mathrm { t } ; \mathrm { y } ( 0 ) = 0y′=tcost;y(0)=0

Given the Differential Equation with the Given Initial Condition $\mathrm { y } ^ { \prime } = \mathrm { t } \cos \mathrm { t } ; \mathrm { y } ( 0 ) = 0$