# What is a particular solution to the differential equation #(du)/dt=(2t+sec^2t)/(2u)# and #u(0)=-5#?

applying the IV

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Grouping variables

Integrating both sides

but considering the initial conditions

and finally

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To find the particular solution to the given differential equation ( \frac{du}{dt} = \frac{2t + \sec^2 t}{2u} ) with the initial condition ( u(0) = -5 ), we need to solve the differential equation.

First, separate variables and integrate both sides with respect to ( u ) and ( t ):

[ \int 2u , du = \int (2t + \sec^2 t) , dt ]

This gives us:

[ u^2 = t^2 + \tan t + C ]

To find the value of the constant ( C ), we use the initial condition ( u(0) = -5 ):

[ (-5)^2 = 0^2 + \tan(0) + C ]

[ 25 = 0 + 0 + C ]

[ C = 25 ]

So, the particular solution to the differential equation is:

[ u^2 = t^2 + \tan t + 25 ]

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