y f(x, y) = x²e" -1 3 1.3 2.1 3.

Q: 2 dydx + f(x, y)dydx

A:

Q: H2. 2dy dy dr2 d²y dy :1 at x = 0. dx – 52+2y= 2x², y=7 and - dx

A:

Q: 3. II Azy'dz dy = 2y°(3² – 1)dy | 2y (y? – 1)dy 1

A: Given that a double integration .We use general  property of integration to solve this integration.…

Q: Solve Separable (x – 4)y*dx – x³ (y² – 3)dy = 0 -

A:

Q: 1. (x- v)dx+ (3x + y)dy = 0 when x = 2, y = -1

A: Let us first solve x-ydx+3x+ydy=0. So,            x-ydx+3x+ydy=0Let y be the dependent variable.…

Q: 8. Jydx + (1+ x)dy = 0,,when x = 1, y = 4

A: Differentiable equation can be solved separating the variables on each side and then integrating…

Q: 22 2 dy = 1-x +y _xy 2 2 x y dx

A:

Q: solve the differenhal equa hon by varianon of parameters y" + 3y't 2y 7 te* y (x)

A:

Q: 8. Find dy dx' x+2x given y = u³ + 4, u = .2

A:

Q: 1 [ (2 -y - VE) dy 2 – y – 0. 1 | (2-y (2 – y – V9) dy 0, 2 | (x² – (2 – æ)) dæ - 2 [ (2² – (2 – a))…

A:

Q: (x²-xy-y²\ (x2-ху-у?) dy = 0 dx + r2-v3

A:

Q: dy/dt^2 – 4dy/dt – 5y =0, y(1) =0 ; y’(1) =2

A: Given differential equation d2ydt2-4dydt-5y=0  , y(1)=0 and y'(1)=2

Q: I-x² f (x, y, 2) dz dy dx 1-x² Jo

A: we have to find the region of integration

Q: dy 4 --v = -r dx 4 P(x) Q(x) = -x³, Linear in 'v'

A: The given equation is: dydx-4xv=-x3px=-4x, qx=-x3

Q: Find aff dy dx² for f(x, y) = exy

A: I have done solution in detail....step by step....go through the solution

Q: 2x(y + 1) dx – ydy = 0 when x = 0, y = -2 %3D -

A: Here differential equation is given as 2xy+1dx-ydy=0or2xy+1dx=ydy⇒2xdx=ydyy+1⇒2xdx=dy-dyy+1

Q: dy 5. S: y(1-Vy)2

A:

Q: dy y dx 2x = 3y2 – 2 when y = 1 and x = 1 in terms of x. %3D - -

A:

Q: 2 In sec Vi + tag V + C sec sec z dz %D

A: 35. given integral,∫sec x . dxx = 2∫sec x12 . 12x-12dxlet x12 = u⇒ 12x12-1 dx = du⇒12x-12dx = du

Q: 3y 6. S dy y-3 7. SVIV7+xVx dx 8. S dr V1+r3

A:

Q: dy 21+ dr2 (vii) dy\2] & (dy\- da2 \dx². (viii) = 0. (ix) () z\ip, + 5y3 = x. dx.

A: (vii) The equation given is d2ydx2=21+dydx232. Squaring both sides. d2ydx22=41+dydx23 Order = 2…

Q: (x³ + y² + z) dx dy dz. 2 Evaluate

A: To evaluate the integral

Q: Obtain the general solution using variation of parameter. (D² + 1)y= cotx

A:

Q: (2y cosx + sin* x)dx sin x dy, when x = y = =,y= 1

A: Given - 2ycosx + sin3xdx = sinx dy , when x = π2, y = 1 To find - Solve the following differential…

Q: 3z cos(z)dz

A:

Q: y 3D 1 y = secxtanx; y

A:

Q: dy of y=X Find

A: y = x^x Find dy/dx

Q: 3. fz √z dz 4. (2x² 3x + 5) dx -

A: 3. ∫zzdz4. ∫2x2-3x+5dx

Q: Evaluate (2x - y) dx + (x + 3y) dy. C: x-axis from (0, 0) to (0, -3) and (0, -3) to (8, -3)

A:

Q: dy اذا كانت y = 3z2 + (sin(z وكان = *3ez؟ 4 : أحسب dx

A:

Q: Z eXZ 4. xyz dydx dz

A: Solution:

Q: dy dz 2. x+1 y+1 z+1

A:

Q: Evaluate the integrals

A: Consider the given integral, ∫02∫-4-y24-y2∫02x+ydz dx dy integrate with the given order of the…

Q: 26. x = 2y, y(1) = 5 dy

A:

Q: b. (xy - y)dx + (x² – 2x + 3y)dy = 0; when x = 1, y = 1

A:

Q: 2x - y + 5) dx - (2x + y) dy = 0 where x= 3, y=7

A: Let's find.

Q: 2. x3 – y3 = c³, find y" %3D d²x 3. x* = at?, find dt2

A: we need to find d2ydx2 from equation x3-y3=c3

Q: dy dy + 7 dt + 10y = e-t dt2

A:

Q: What is the I.F of (x^2y^3+2y)dx+(2x-2x^3y^2)dy=0?

A: Find the integrating factor

Q: 3. Integrate! 12e-0.21 а) -dt b) 6r 3,2 Se041 dr dx c) f- x. V1+ In 3x?

A:

Q: IIT /2 pcos x ex-X sin(x + y)dy dz dx Jz /6 Jo

A: Given integration is: ∫π6π2∫0cosx∫xπ-xsinx+ydxdydz  To find the value of integration.

Q: Hei! If y cos(xy) dydx = x=0 y=0

A:

Q: Sx² cos (x3 + 3) Van (x3• 3) dx + 3) Vsin (x3+ 3) dx 1)

A:

Q: 1 dz -÷z³ = t² + 3t – 2 dt 3 z2 -

A: Differential equations and their type

Q: cos z dz ar gral f

A:

Q: Evaluate (2x - y) dx + (x + 3y) dy. C: x-axis from (0, 0) to (0, -3) and (0, -3) to (4, -3)

A: To solve the given line integral.