8:43 B O O * all all 73%t cln 10 Evaluate In 8 -dx et – 5

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A: since you have asked about only part a  Given initial value problem dydt=yt3-1.5y     ; y(0)=1

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A: Please see the below picture for detailed solution. 

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A: Given that: ∫dxx+x3

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A:

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A: Given that ∫033x-2x2+2x+17dxSolution : put 3x-2=Addxx2+2x+17+B3x-2=A2x+2+Bput x=-13-1-2=A-2+2+BB=-5p...

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A: Given:       dy-ylnyecosxdx=ysinxdx

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A: We can find the integral as below.

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A: option D

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A: Topic = Derivative

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A: The nth term is given below

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A: Solution

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A: Explanation of the answer is as follows

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A:

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A: Let I=∫1xx6-1dxPut x6-1=u(squaring on both sidex6-1=u2x6=u2+1)differentiate 12x6-1ddx(x6-1)=du12x6-1...

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A: (x+2)y' = xy + 2y - 3x - 6  So, (x+2)y' = (x+2)y - 3(x+2)  So, (x+2)y' = (x+2)(y-3) 

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A: Solution