**Unformatted text preview: **y ( t ) =-t 2 / 4, which is another solution of the diﬀerential equation. If y ( t ) = ct + c 2 then y ( t ) = c . On the other hand-t + ( t 2 + 4 y ( t )) 1 / 2 2 =-t + ( t 2 + 4 ct + 4 c 2 ) 1 / 2 2 =-t + ( t + 2 c ) 2 = c. So, no matter what c is equal to, y ( t ) is a solution. It should be clear that there is no c such that ct + c 2 =-t 2 / 4 since this would mean that, for all t , 1 4 t 2 + ct + c 2 = 0, which can’t happen. 1...

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