Q. 37 F5.0( 1 Vote )

# Solve each of the following initial value problems:, y(0) = 1

, 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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