(D² + 1)y = sec³ x.

Q: dy y² tion + dx x² है || Y 8

A:

Q: Find the solution fo (x^3-y^3)dx + xy^2

A: Here we have,  (x3-y3)dx+xy2dy=0⇒(x3-y3)dx=xy2dy⇒dydx=(x3-y3)xy2 ..............(1) Which is a…

Q: 3. f 3/2 y dydx (I wz dx dy dz. 4.

A:

Q: Find d2y/dx2 in terms of x and y. x2y2-10x=1

A:

Q: Solve the homogenous diff equation of; y' = y sec x

A:

Q: (f) / sec“ y dy. tan5

A:

Q: (») LLa (1-y? In(x² + y² + 1) dx dy 1-у2

A: Given: ∫-11∫-1-y21-y2    lnx2+y2+1dxdy

Q: Solve for dy y = sec 2x² dx

A:

Q: Find dx y = 9x -2 dy dx II

A: Given:    y = 9x-2

Q: dy ニ 3 x +y-2

A:

Q: If xy + y2 = 1, find the value of d2y/dx2 at the point (0, -1).

A:

Q: d'y 4x dx2 dy +6- + у %3D0 dx

A:

Q: sec?xsec'y dx dy .2

A:

Q: dy y = (3x2 + x-2)-1; dx

A:

Q: dy - 2rV1 – y³, -1 < y< 1 dx

A: Given dydx=2x1-y2

Q: Evaluate the integrals

A: The given integral is ∫02∫03y∫x2+3y28-x2-y2dzdxdy.

Q: calculate d2y/dx2 when y=3/x

A:

Q: d²y Find xp y = d²y 口 %3D dx 2.

A: topic - double derivative of a function

Q: Find d2y/dx2 implicitly in terms of x and y.

A: To findd2ydx2

Q: dy dx2 in terms of x and y. Find x2y2 - 4x = 3 dy dx2

A: Given equation :

Q: y=(sinx)x    d2y/dx2 ?

A: Apply exponent rule and solve:

Q: Y = 4x²-5 Vsx dy=?

A: We want to find derivative of y, The given expression for y is,               y=4x5-55x

Q: 3. (D2 + 2D + 1)y = 0; when x = 0, y = 1, Dy = - 1.

A:

Q: (x² + y2) dx dy ow urves y = x² and

A:

Q: y = ( 2x³ - 8x )* Find dy/dx when x = -%

A:

Q: calculate d2y/dx2 when y=0.4x-0.3

A: use this formula:

Q: colve dx+. y 2х y²-3x² -dy=0 y?

A: Given 2xy3dx+y2-3x2y2dy=0Need to solve

Q: find in terms of x and y. dx 18) x =-4y+2y

A: x2 = -4y +2y3 To find dy/dx in terms of x and y

Q: dy. dx2 y = -6x2 + 7x Calculate d²y dx2

A:

Q: d²y + y = 0 dx2

A:

Q: Q of y = e(cosx+4tanx+5t-tx²) dp

A: We have to find derivative of y with respect p.

Q: -2x (x² + y²)dydx J.

A: ∫02∫x22x(x2+y2) dydx

Q: Q9// find the general solution (sec³x + (cotx)y)dx + (Insinx)dy = 0

A: Answer and explanation is given below...

Q: dy un 3 sec2 5x, w dr

A:

Q: Let y = 4**-2x+3. Find d²y - dx2°

A:

Q: Find and in terms of t. dx2 x= -9 sint, y = cost dx dx2

A:

Q: y = e² -Vx sec 6x 3

A:

Q: COs 2 dydx (4) LL. r1 1 dydx x4 + y²

A:

Q: if x3 + y3 =65, find the value of d2y/dx2 at the point (4,1)

A: Given that,    x3+y3=65

Q: Solve: y(x) dy dx = = (8x² - 7x + 6) (y-7), where y(0) = 11.

A: We will solve the given differetial equation by variable separation method.

Q: calculate d2y/dx2 when y=e-(x-3)-x

A:

Q: calculate d2y/dx2 when y=-2/x2

A: Given: y=-2x2

Q: y = (x2 - 4)2(5 - 3x). Find dy/dx when x = 3.

A: We have to find dydx at x=3 for the function: y=x2-425-3x Since two function are in product so we…

Q: Find DF of fcx,y)=x²e° at the point (-2,0)-

A:

Q: d²y Find dy 34 y3 %3D こX+

A: Answer

Q: y 2t3-5t2 +3, find dy/dt

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

Q: dy 3 y : 2 dx 1 sec2 y || + II

A:

Optimization

Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the selection of the best element for a given system.

Integration

Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.

Application of Integration

In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.

Volume

In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects that have length, breadth, and height (or depth).

Area

Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.