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Exercise

dydx=(1−y)(10−y)\frac{dy}{dx}=\left(1-y\right)\left(10-y\right)

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

1(1−y)(10−y)dy=dx\frac{1}{\left(1-y\right)\left(10-y\right)}dy=dx
2

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

∫1(1−y)(10−y)dy=∫1dx\int\frac{1}{\left(1-y\right)\left(10-y\right)}dy=\int1dx
3

Solve the integral ∫1(1−y)(10−y)dy\int\frac{1}{\left(1-y\right)\left(10-y\right)}dy and replace the result in the differential equation

−19ln⁡∣−y+1∣+19ln⁡∣−y+10∣=∫1dx-\frac{1}{9}\ln\left|-y+1\right|+\frac{1}{9}\ln\left|-y+10\right|=\int1dx
4

Solve the integral ∫1dx\int1dx and replace the result in the differential equation

−19ln⁡∣−y+1∣+19ln⁡∣−y+10∣=x+C0-\frac{1}{9}\ln\left|-y+1\right|+\frac{1}{9}\ln\left|-y+10\right|=x+C_0

Final answer to the exercise

−19ln⁡∣−y+1∣+19ln⁡∣−y+10∣=x+C0-\frac{1}{9}\ln\left|-y+1\right|+\frac{1}{9}\ln\left|-y+10\right|=x+C_0

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
  • Integrate by substitution
  • Integrate by parts
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dydx =(1−y)(10−y)
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