Solve: x- dx dy - (x + 1)(1 + y*).

Q: use appropriate substitution to solve differential equatio dy/dx-(2/x)y=-x^2y^3

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Q: solve using Separation of Variables. dy dt = ry-3

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Q: Find dy/dx for (x+y)3=x3+y3,at the point  (-1,1)

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Q: Find dy/dx by implicit differentiation. 8x3 + x?y – xy3 = -8x2 y' = 2/3

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Q: Integrate:  (5e^-3x - cos (6x))dx

A: Integrate

Q: dx S COsxtanx

A: Here we will solve the given integration,

Q: find dy/dx of y=cos2x-sin4x

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Q: Solve (Vr) = sec (Vx) using variable separable form. dy dx

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Q: EXERCISE 1 dy -dx = 0

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Q: x3sec2y dy/dx+3x2tan y=cos

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Q: Find dy/dx at the point (2,1) if 4y3-x2y2=0

A: Given that: 4y3-x2y2=0

Q: tanx dx = -In/secx|+C^O .BO In|secx|+CO In/cscx| +CO -In|cscx|+C°O .D

A: We need to evaluate the following integral I = ∫tanxdx

Q: = vt – 2, d²y t – 2. d²y y = t² – 3' dx2

A: x=t-2y=t2-3

Q: (solve exact de) (2ysinxcosx - y + 2y²e*v¨ )dx = (x - sin'x - 4xye )dy

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Q: /4 - x2 the integral | dx, becomes

A: Given the integrationI=∫4−x2x2dx

Q: find dy / dx by implicit differentiation.

A: Given equation is x4y-8xy+3xy2=9. To find dydx. Solution: Power rule of differentiation states that:…

Q: arctanx e dx (1+x²,"

A: The given problem is to evaluate the given integral. We have to evaluate by using substitution…

Q: Use differentiation rules to find y' = dy/dx.

A: Rearrange y to get y in fractional form.

Q: Find dy/dx by implicit differentiation. 3x2 - 3y2 = 16 %3D dy/dx =

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Q: Differentiate. y =x° In x- dy II

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Q: find dy / dx by implicit differentiation.

A: We have use implicit differentiation in the following equation to find dydx. 2x3 + 3y3 = 64

Q: Find dy/dx at the point (2,-2) if 3y2-4x2+4=0.

A: Find dy/dx at the point (2,-2) if 3y2-4x2+4=0.

Q: 5. Solve dy dy + y = 2sin(3x) dx - dx? using the method of undetermined coefficient.

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Q: Find dy/dx by implicity differation.

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Q: 7) y = exz 8) y = 3 x ²

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Q: (a) --y = 2x?y, y(1) = 1 dy dx

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Q: (continue solution of de) (2ysinxcosx - y + 2y²e*Y" )dx = (x - sin²x - 4xye*y" )dy

A: Using method of exact differential equation.

Q: Find dy/dx at the given point 6x2-5y3=2x-5y at (0,1)

A: First differentiate the equation and then substitute the value of x as 0 and y as 1.

Q: dx Q2. Integrate

A: Given,I=∫1x2+15dx Consider x=tanu and u=arctan(x) then, dx=sec2udu…

Q: Solve (1 + y) dx = (tanly-x) dy

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Q: 2х — 1 - dx 6х + 18 x² -

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Q: In integrating the function fx cos(3x) dx using integration by parts, what would you choose for your…

A: Let ,I=∫ xcos3xdxSubstitute  u=3xdifferentiate du=3dx⇒du3=dxI=∫ u3cosudu3I=∫ucosu9duTake the…

Q: y= sinh-lx then prove that Vx2+1 dx

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Q: solve using homogenous and exact. (2x-3y)dx+(2y-3x)dy=0

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Q: dy Exc:" of y = tanh (x5 – 5x)² --

A: Solution:Given         y=tanh4(x5-5x)2  To find dydxFormula:ddx(xn)=nxn-1ddx(tanh x)=sech2x

Q: Find dy/dx by implicit differeniation x2y+y2x =- 2

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Q: dy Integrate dx = 4e®xand find equation of curve that passes through the point (1,7).

A: Given that, dydx = 4e8x

Q: Solve dy dy + y = 2sin(3x) - dx? dx using the method of undetermined coefficient.

A: We solve the given differential equation by method of undetermined coefficients.

Q: Integrate both sides dy = -dx

A: the given integral equation is                             ∫1y2dy=∫1xdx

Q: dy dx 118 (x = 3x - 5

A: Topic:- differential equations

Q: dy Use an appropriate technique to find dx from x² + y² = (2x² + 2y² -x)*.

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Q: Integrate dy/dx=(1+x^2)/(-2x-x^3)

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Q: 3. (2х - у - 4) dx - (х-3)dy -0 when x %3 1, y 3D 4.

A: We try to identify and group terms and integrate.

Q: Find dy/dx by implicit differentiation. бх3 + x2у - хуз %3D9 y' = %3D

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Q: . Use separable equations to find the general and particular solution of xy dx – (x +

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Q: Cosx 2) ). y dx+ týž - Sinx (²) dy = уч

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Q: 1 dy if y = dx -1 Find - tan 3x. dy dax

A: Use derivative formula

Q: The second derivative d^2y/dx^2 of y^3 = xy + 1 at the point (0,1) is

A:

Q: Obtain the particular solution. 1. 2y dx = 3x dy ; when x = 2 , y = -1 .

A:

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.