Given the differential equation: $ty ^ { \prime } = \ln t$ , is this the solution $y = \frac { ( \ln t ) ^ { 2 } } { 2 } + C ?$

Question

Quizplus · Accepted Answer

To verify if $y = \frac { ( \ln t ) ^ { 2 } } { 2 } + C$ is a solution to the differential equation $ty' = \ln t$, differentiate $y$ with respect to $t$. The derivative $y'$ is $\frac{d}{dt}\left(\frac { ( \ln t ) ^ { 2 } } { 2 } + C\right) = \frac{1}{t} \ln t$. Multiplying both sides by $t$ gives $ty' = \ln t$, which matches the given differential equation.

Given the Differential Equation ty′=ln⁡tty ^ { \prime } = \ln tty′=lnt

Given the Differential Equation $ty ^ { \prime } = \ln t$