Q. 37 F5.0( 1 Vote )

Solve each of the following initial value problems:

, y(0) = 1

Answer :

, y(0) = 1


Given and y(0) = 1



This is a first order linear differential equation of the form



Here, P = tan x and Q = 2x + x2tan x


The integrating factor (I.F) of this differential equation is,




We have



I.F = sec x [ elog x = x]


Hence, the solution of the differential equation is,








Recall







y sec x = x2sec x + c




y = x2 + c cos x


However, when x = 0, we have y = 1


1 = 02 + c(cos 0)


1 = 0 + c(1)


c = 1


By substituting the value of c in the equation for y, we get


y = x2 + (1)cos x


y = x2 + cos x


Thus, the solution of the given initial value problem is y = x2 + cos x


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