Question

A.) Evaluate: Integration of (3t+t^2﻿﻿﻿) sin(2t)dt
B.) Solve: {2xy cosx^2 - 2xy + 1}dx + {sinx^2 - x^2 + 3}dy = 0

Accepted Answer

* Integration of (3t+t^2﻿﻿﻿) sin(2t)dt

((-t2cos(2t))/2)+((_t_sin(2_t_))/2)-((3t cos(2t))/2+((cos2t))/4
+(3sin(2t))/4 + C

2 .

(1+2xy cosx2 -2xy +(sinx2 -x2 )dy/dx=0

(1+2xy cos(x2 ) -2xy +(sinx2 -(x2 ) )y=0

rewrite in the form of a first order linear ODE

y + (2x(cos(x2 )-1))/(-x2 + sin (x2))y = -1/sin(x2)-x2

integration factor = -x2+sin(x2)

((-x2+sin(x2))y)=-1

y=-(x /( -x2 +sin(x2))+c1/( -x2 +sin(x2))

Question #229205

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