Given the differential equation: $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { \mathrm { t } + \mathrm { e } ^ { \mathrm { t } } } { \mathrm { y } } \text {, }$ is this the solution $y = \pm \sqrt { t ^ { 2 } + 2 e ^ { t } + C } ?$

Question

Quizplus · Accepted Answer

To verify if $y = \pm \sqrt { t ^ { 2 } + 2 e ^ { t } + C }$ is a solution to the given differential equation $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { \mathrm { t } + \mathrm { e } ^ { \mathrm { t } } } { \mathrm { y } }$, we differentiate $y$ with respect to $t$ and substitute back into the original equation. Differentiating $y$ gives $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac {1}{2} \left( t ^ { 2 } + 2 e ^ { t } + C \right)^{-1/2} \cdot (2t + 2e^t)$, which simplifies to $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { t + e ^ { t } } { \sqrt { t ^ { 2 } + 2 e ^ { t } + C } }$. This matches the given differential equation when $y = \pm \sqrt { t ^ { 2 } + 2 e ^ { t } + C }$, confirming the solution is correct.

Given the Differential Equation dydt=t+ety, \frac { \mathrm { dy } } { \mathrm { dt } } = \frac { \mathrm { t } + \mathrm { e } ^ { \mathrm { t } } } { \mathrm { y } } \text {, }dtdy​=yt+et​,

Given the Differential Equation $\frac { \mathrm { dy } } { \mathrm { dt } } = \frac { \mathrm { t } + \mathrm { e } ^ { \mathrm { t } } } { \mathrm { y } } \text {, }$