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Exercise

dydx=1−2cosx\frac{dy}{dx}=1-2cosx

Step-by-step Solution

1

Group the terms of the differential equation. Move the terms of the yy variable to the left side, and the terms of the xx variable to the right side of the equality

dy=(1−2cos⁡(x))dxdy=\left(1-2\cos\left(x\right)\right)dx
2

Integrate both sides of the differential equation, the left side with respect to yy, and the right side with respect to xx

∫1dy=∫(1−2cos⁡(x))dx\int1dy=\int\left(1-2\cos\left(x\right)\right)dx
3

Expand the integral ∫(1−2cos⁡(x))dx\int\left(1-2\cos\left(x\right)\right)dx into 22 integrals using the sum rule for integrals, to then solve each integral separately

∫1dy=∫1dx+∫−2cos⁡(x)dx\int1dy=\int1dx+\int-2\cos\left(x\right)dx
4

Solve the integral ∫1dy\int1dy and replace the result in the differential equation

y=∫1dx+∫−2cos⁡(x)dxy=\int1dx+\int-2\cos\left(x\right)dx
5

Solve the integral ∫1dx+∫−2cos⁡(x)dx\int1dx+\int-2\cos\left(x\right)dx and replace the result in the differential equation

y=x−2sin⁡(x)+C0y=x-2\sin\left(x\right)+C_0

Final answer to the exercise

y=x−2sin⁡(x)+C0y=x-2\sin\left(x\right)+C_0

Try other ways to solve this exercise

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  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equations
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Integrate by parts
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Similar Questions Solved

dydx=y2−9\frac{dy}{dx}=y^2-9

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dydx=y−y2\frac{dy}{dx}=y-y^2

dydx=y(y+2)\frac{dy}{dx}=y\left(y+2\right)

dydx=(1+e−x)(y2−1)\frac{dy}{dx}=\left(1+e^{-x}\right)\left(y^2-1\right)

dydx=x2x−y\frac{dy}{dx}=\frac{x}{2x-y}

dydx=y(1−y)\frac{dy}{dx}=y\left(1-y\right)

Main Topic: Separable Differential Equations

A separable differential equation is one that can be solved by separating variables, that is, when all expressions involving a variable can be placed on one side of the equation, and expressions involving the other variable on the other side of the equation.

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