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Exercise

sinθ −sin(3)θ =sinθ cos(2)θ sin\theta\:-sin^{\left(3\right)}\theta\:=sin\theta\:cos^{\left(2\right)}\theta\:

Step-by-step Solution

1

Rewrite the differential equation using Leibniz notation

sin⁡(θ)−sin⁡(′′θ)′=sin⁡(θ)cos⁡(θ)2\sin\left(\theta\right)- \sin\left(''\theta\right)^{\prime}=\sin\left(\theta\right)\cos\left(\theta\right)^2

Final answer to the exercise

sin⁡(θ)−sin⁡(′′θ)′=sin⁡(θ)cos⁡(θ)2\sin\left(\theta\right)- \sin\left(''\theta\right)^{\prime}=\sin\left(\theta\right)\cos\left(\theta\right)^2

Try other ways to solve this exercise

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  • Exact Differential Equation
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  • Separable Differential Equations
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
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sinθ −sin(3)θ =sinθ cos(2)θ
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