Consider a first order differential equation 2t1+t2y+21+t2

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delilnaT · Accepted Answer

Possible derivation: ddy((2t)y1+t2+21+t2) Differentiate the sum term by term and factor out constants: =ddy(21+t2)+2t(ddy(y))1+t2 The derivative of 2t2+1 is zero: =2t(ddy(y))1+t2+0 INTERMEDIATE STEPS: Possible derivation: ddy(21+t2) The limit definition of the derivative for a function f(y) is ddy(f(y))=limh→0f(y+h)−f(y)h. Let f(y)=21+t2. Then: f(y+h)=21+t2 Substitute f(y + h) and f(y) into the limit definition for the derivative: ddy(21+t2)=limh→021+t2−21+t2h In order to properly evaluate the limit, the h in the denominator needs to be canceled out. Begin by expanding all powers of (y + h): =limh→0−21+t2+(21+t2)h Combine like terms in the numerator and order by descending powers of y: =limh→00h Factor h out of the numerator: =limh→0h(0)h The limit variable h approaches, but is never equal to, 0 and therefore can be algebraically canceled: =limh→00 Evaluate the limit as h goes to 0: =0 Simplify to finish computing the derivative: Answer: ddy(21+t2)=0 Simplify the expression: =2t(ddy(y))1+t2 The derivative of y is 1: Answer: =12t1+t2 INTERMEDIATE STEPS: Possible derivation: ddy(y) The limit definition of the derivative for a function f(y) is ddy(f(y))=limh→0f(y+h)−f(y)h. Let f(y)=y. Then: f(y+h)=(y+h) Substitute f(y + h) and f(y) into the limit definition for the derivative:

Consider a first order differential equation , \frac{{{2}{t}}}{{{1}+{t}^{2}}}{y}+\frac{2}{{{1}+{t}^{2}}}

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