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Exercise

$\frac{dp}{dx}=\frac{-p^2+2p}{p-1}$

Step-by-step Solution

1

Group the terms of the differential equation. Move the terms of the $p$ variable to the left side, and the terms of the $x$ variable to the right side of the equality

$\frac{p-1}{-p^2+2p}dp=dx$
2

Simplify the expression $\frac{p-1}{-p^2+2p}dp$

$\frac{p-1}{p\left(-p+2\right)}dp=dx$
3

Integrate both sides of the differential equation, the left side with respect to $y$, and the right side with respect to $x$

$\int\frac{p-1}{p\left(-p+2\right)}dp=\int1dx$
4

Solve the integral $\int\frac{p-1}{p\left(-p+2\right)}dp$ and replace the result in the differential equation

$-\frac{1}{2}\ln\left|p\right|-\frac{1}{2}\ln\left|-p+2\right|=\int1dx$
5

Solve the integral $\int1dx$ and replace the result in the differential equation

$-\frac{1}{2}\ln\left|p\right|-\frac{1}{2}\ln\left|-p+2\right|=x+C_0$

Final answer to the exercise

$-\frac{1}{2}\ln\left|p\right|-\frac{1}{2}\ln\left|-p+2\right|=x+C_0$

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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