Introduction

Solution integrating both sides gives y x z 1 1 x 2

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Solution Integrating both sides gives y ( x ) = Z 1 1 + x 2 dx = arctan( x ) + C.

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Initial Value Problems An initial value problem (IVP) is an n th-order differential equation together with enough conditions of the form y ( x 0 ) = y 0 , y 0 ( x 0 ) = y 1 , y 00 ( x 0 ) = y 2 , . . . so that all of the arbitrary constants in the general solution to the differential equation are deter- mined. In general, a n th order differential equation will require n such conditions. The solution to an IVP is called a specific solution to the differential equation. I Example Find a solution to the initial value problem y 0 = 1 1 + x 2 , y (1) = 2 . Solution We have already seen that the general solution is y ( x ) = arctan( x ) + C. Plugging in x 0 = 1 and y ( x 0 ) = 2 we have 2 = arctan(1) + C = π/ 4 + C so C = 2 - π/ 4 . Then the solution to the IVP is y ( x ) = arctan( x ) + 2 - π/ 4 .
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