Solve the ODE: dy (x' +4x) 0 using separable variable method: dx x(sin 2y +4)

Q: 3x4-4 cos x = -6

A: Find d2y/dx2 implicitly in terms of x and y3xy − 4 cos x = −6

Q: (xy^2+y)dx -xdy =0

A: Given: (xy2+y)dx-x dy=0

Q: Solve Initial value problem i. 8cos?ydx + cosec²xdy=0, y() = dy _ x(r+1) 4y3 y(0) =- 1/V2 ii. dx

A: "Since you have asked multiple question, we will solve the first question for you.If you want any…

Q: sinhx dx 1+ coshx ) A 1+ coshx B (1+coshx)? (1+coshx)² D 1+ coshx

A: We have to find

Q: 3. (y-x)dx = xy (x dx + y dy)

A:

Q: solve xsinydx+(x2+1)cosydy=0

A: Given differential equation is: x siny dx+(x2+1) cosy dy=0 Use variable separable method to solve…

Q: 8 him XT 2 (secx tanx) 1

A:

Q: (ど4ー) 2dx 4 X(3(Inx)*+121nx-15)2 2)

A:

Q: Simplify 3π2 sinh (πx) - π2 cosh (πx) - 3 sinh (πx) + cosh (πx) = 0 on the interval (0,1).

A:

Q: sec? (1/x) dx. 9: (16pts) Evaluate

A:

Q: differentiate y = (x - 1)/(x + 1)

A: Given function: y=x-1x+1 Differentiate the above function w.r.t 'x' dydx=ddxx-1x+1

Q: (coshx – sinhx)? dx =

A: To evaluate the given integral.

Q: Ssin8 (3z) cos5 (3z) dz

A:

Q: dr(sec'x) dx

A:

Q: solve the ODE : xydx + ( x^2 + y^2 + 1 ) dy = 0

A: Solution: The objective is to find the general solution of the given differential equation

Q: y= Sinh (3x+5x+6) 2. %3D

A: We have to find differentiation of the given function.

Q: The question is:  y=(x^2+x^3)^4 , find y’

A: The given equation is y =(x^2+ x^3)^4.By the chain rule of differentiation, compute y'.

Q: d^2y/dx^2 + 3dy/dx + 4y = 3x + 2 Utilize the indeterminate coefficient technique to solve the above…

A: Here is the, solution

Q: (4 - x2)3/2 dx 3

A: Consider, I=423∫-22(4-x2)3/2dx I =4232∫02(4-x2)3/2 dx   since f(x) is even function that is…

Q: Differentiate: y= tan(8x2-3x*) %3D O y' = (16x - 12a) sec(8a? - 3x*) tan(8r2- 3r) O y' = (16x – 12a)…

A: Given; y = tan(8x2-3x4)