(2 + y²) dx dy 0,

Q: -ydx + (x + √√xy) dy=0

A: answer is in next step; please give a like !!

Q: (2+3x)2 dx

A:

Q: 2x-1 dx

A: Given,

Q: 2. メ+ 2x

A: I have given an answer in step 2.

Q: (6x + y2) dx + y(2x - 3y) dy = 0. evaluate

A: The given differential equation is                                    (6x+y2)dx+y(2x-3y)=0

Q: V1+e-z dr.wo ez

A:

Q: Solve, (2x+3y-2)dx+(3x-4y+1)dy = 0 %3D

A:

Q: –dx (1 +x) (x+1)Vx² +2x-2

A:

Q: dy dx 2 x +34-4 4メ+y -3

A:

Q: Solve 2- (2xy - 2y)dx + (3x'y-4x)dy 0

A: We have to solve given problem:

Q: (2x + 1) dx

A: To evaluate the integral: ∫02(2x2+1)dx. Solution: ∫02(2x2+1)dx On simplifying further we get:…

Q: dy + (1x – 7)y = 2x7 dx

A:

Q: 2y d³y dx3 (y + 3 dx) dy dx² + 2

A:

Q: dx 1. dy y 3x :-5x2 dy 2y 2. dx = -4x²

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Q: 2r-r2 -3 zVx2 + y2 dz dy dx. 0. 0.

A:

Q: xdy = (y - Vx² + y²) dx

A:

Q: | (3 + xx) dx

A:

Q: L Lyx² exydxdy -y

A: We need to evaluate ∫-10∫-y1x2 exy dx dy The region of integration is given as Here -y≤x≤1-1≤y≤0 On…

Q: Show that (xcos(2x)dx=. A-2 8

A: LetI=∫0π4xcos2xdxPut 2x=udifferentiate on both side2dx=dudx=du2As x→0 then u→0As x→π4 then…

Q: -dx x +1 2 5.

A:

Q: (1 +Iпx) dx x3

A: The given integral is ∫1+ln xx3dx. Here, u=1+ln x  and  v=1x3.

Q: 1 = f(y-x(2-x))dx 21

A: Given: I=∫02δy-x2-x dxTo find: The value of the given integral.

Q: dy dx x3 2+y²

A:

Q: Y(Gyt-x-4)dx + 2x dy =0

A: let

Q: Solve (2xy-2xy- = 0 4xy² +2x) dx + (3x2y + 4y) dy %3D

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Q: dy Find if dx y = (3x2 + 6x – vx)(5x + + 10). DO NOT SIMPLIFY.

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Q: sec2 ) 15. Evaluate. - x2 dx

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Q: Find 9 dx

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Q: Find d2y/dx2. y=5x3/3   -2x

A: