1. Bernoulli Equations
A Bernoulli equation has the form:
\frac{dy}{dx} + P(x)y = Q(x)y^n To solve,
Divide by y^n \.
Make the substitution v = y^{1-n} \.
Transforming it into a linear equation.
Solve the linear equation.
Resubstitute back.
2. Examples
Example 1:
Solve
Solution:
1. Divide by y^2 \:
\frac{1}{y^2} \frac{dy}{dx} + \frac{1}{y} = 1\
2. Substitute v = y^{1-2} = \frac{1}{y} \:
\frac{dv}{dx} = -\frac{1}{y^2} \frac{dy}{dx}\
Thus,
-\frac{dv}{dx} + v = 1 \quad \Rightarrow \frac{dv}{dx} - v = -1
3. Solve the linear equation:
- Integrating factor:
\mu(x) = e^{\int -1 \, dx} = e^{-x}
- Multiply through by e^{-x}:
e^{-x}\frac{dv}{dx} - e^{-x}v = -e^{-x}
- Recognize the left side as a derivative:
\frac{d}{dx} (ve^{-x}) = -e^{-x}
- Integrate both sides:
ve^{-x} = \int -e^{-x} \, dx \quad \Rightarrow ve^{-x} = e^{-x} + C
- Solve for v:
v = 1 + Ce^x
4. Substitute back v = \frac{1}{y} \:
\frac{1}{y} = 1 + Ce^x
5. Final solution:
y = \frac{1}{1 + Ce^x} \
Example 2:
Solve
\frac{dy}{dx} + 2y = x y^2 Solution:
1. Divide by y^2 \:
\frac{1}{y^2} \frac{dy}{dx} + \frac{2}{y} = x\
2. Substitute v = y^{1-2} = \frac{1}{y} \:
\frac{dv}{dx} = -\frac{1}{y^2} \frac{dy}{dx}\
Thus,
-\frac{dv}{dx} + 2v = x \quad \Rightarrow \frac{dv}{dx} - 2v = -x
3. Solve the linear equation:
- Integrating factor:
\mu(x) = e^{\int -2 \, dx} = e^{-2x} \
- Multiply through by e^{-2x}:
e^{-2x} \frac{dv}{dx} - 2e^{-2x}v = -xe^{-2x}
- Recognize the left side as a derivative:
\frac{d}{dx} (ve^{-2x}) = -xe^{-2x}
- Integrate both sides:
ve^{-2x} = \int -xe^{-2x} \, dx
Use integration by parts: u = x, dv = -e^{-2x} dx, then du = dx \, v = \frac{1}{2} e^{-2x}:
\int -xe^{-2x} \, dx = \frac{x}{2} e^{-2x} - \int \frac{1}{2} e^{-2x} \, dx = \frac{x}{2} e^{-2x} + \frac{1}{4} e^{-2x} + C
- Solve for v:
ve^{-2x} = \frac{x}{2} e^{-2x} + \frac{1}{4} e^{-2x} + C
v = \frac{x}{2} + \frac{1}{4} + Ce^{2x}
4. Substitute back v = \frac{1}{y} \:
\frac{1}{y} = \frac{x}{2} + \frac{1}{4} + Ce^{-2x}
5. Final solution:
y = \frac{1}{\frac{x}{2} + \frac{1}{4} + Ce^{-2x}}
Example 3:
Example 3:
Solve
\frac{dy}{dx} + \frac{2y}{x} = x^2 y^3 Answer
Solution:
1. Divide by y^3 \:
\frac{1}{y^3} \frac{dy}{dx} + \frac{2}{xy^2} = x^2\
2. Substitute v = y^{1-3} = \frac{1}{y^2} \:
\frac{dv}{dx} = -\frac{2}{y^3} \frac{dy}{dx} \quad \Rightarrow \frac{1}{y^3} \frac{dy}{dx} = -\frac{1}{2} \frac{dv}{dx}
Thus,
-\frac{1}{2} \frac{dv}{dx} + \frac{2}{x} v = x^2 \quad \Rightarrow \frac{dv}{dx} - \frac{4}{x} v = -2x^2
3. Solve the linear equation:
- Integrating factor:
\mu(x) = e^{\int -\frac{4}{x} \, dx} = e^{-4\ln |x|} = \frac{1}{x^4}
- Multiply through by \frac{1}{x^4}:
\frac{1}{x^4} \frac{dv}{dx} - \frac{4}{x^5} v = -\frac{2}{x^2}
- Recognize the left side as a derivative:
\frac{d}{dx} \left(\frac{1}{x^4} v\right) = -\frac{2}{x^2}
- Integrate both sides:
\frac{1}{x^4} v = \int -\frac{2}{x^2} \, dx = \frac{2}{x} + C
- Solve for v:
v = 2x^3 + Cx^4 \
4. Substitute back v = \frac{1}{y^2} \:
\frac{1}{y^2} = 2x^3 + Cx^4
5. Final solution:
y^2 = \frac{1}{2x^3 + Cx^4}
Example 4:
Example 4:
Solve
\frac{dy}{dx} + \frac{3y}{x} = \frac{y^2}{x^2} Answer
Solution:
1. Divide by y^2 \:
\frac{1}{y^2} \frac{dy}{dx} + \frac{3}{xy} = \frac{1}{x^2}\
2. Substitute v = y^{1-2} = \frac{1}{y} \:
\frac{dv}{dx} = -\frac{1}{y^2} \frac{dy}{dx} \quad \Rightarrow \frac{1}{y^2} \frac{dy}{dx} = -\frac{dv}{dx}
Thus,
-\frac{dv}{dx} + \frac{3}{x}v = \frac{1}{x^2} \quad \Rightarrow \frac{dv}{dx} - \frac{3}{x}v = -\frac{1}{x^2}
3. Solve the linear equation:
- Integrating factor:
\mu(x) = e^{\int -\frac{3}{x} \, dx} = e^{-3 \ln |x|} = \frac{1}{x^3}
- Multiply through by \frac{1}{x^3}:
\frac{1}{x^3} \frac{dv}{dx} - \frac{3}{x^4}v = -\frac{1}{x^5}
- Recognize the left side as a derivative:
\frac{d}{dx} \left(\frac{1}{x^3} v\right) = -\frac{1}{x^5}
- Integrate both sides:
\frac{1}{x^3} v = \int -\frac{1}{x^5} \, dx = \frac{1}{4x^4} + C
- Solve for v:
v = \frac{1}{4x} + Cx^3
4. Substitute back v = \frac{1}{y} \:
\frac{1}{y} = \frac{1}{4x} + Cx^3
5. Final solution:
y = \frac{1}{\frac{1}{4x} + Cx^3}
Example 5:
Example 5:
Solve
\frac{dy}{dx} + \frac{y}{x} = x^2 y^4 Answer
Solution:
1. Divide by y^4 \:
\frac{1}{y^4} \frac{dy}{dx} + \frac{1}{xy^3} = x^2\
2. Substitute v = y^{1-4} = \frac{1}{y^3}:
\frac{dv}{dx} = -\frac{3}{y^4} \frac{dy}{dx} \quad \Rightarrow \frac{1}{y^4} \frac{dy}{dx} = -\frac{1}{3} \frac{dv}{dx}
Thus,
-\frac{1}{3} \frac{dv}{dx} + \frac{1}{x} v = x^2 \quad \Rightarrow \frac{dv}{dx} - \frac{3}{x} v = -3x^2
3. Solve the linear equation:
- Integrating factor:
\mu(x) = e^{\int -\frac{3}{x} \, dx} = e^{-3\ln|x|} = \frac{1}{x^3}
- Multiply through by \frac{1}{x^3}:
\frac{1}{x^3} \frac{dv}{dx} - \frac{3}{x^4} v = -\frac{3}{x}
- Recognize the left side as a derivative:
\frac{d}{dx} \left(\frac{1}{x^3} v\right) = -\frac{3}{x}
- Integrate both sides:
\frac{1}{x^3} v = \int -\frac{3}{x} \, dx = -3 \ln{|x|} + C
- Solve for v:
v = -3x^3 \ln{|x|}+ Cx^3
4. Substitute back v = \frac{1}{y^3} \:
\frac{1}{y^3} = -3x^3 \ln{|x|}+ Cx^3
5. Final solution:
y^3 = \frac{1}{-3x^3 \ln{|x|}+ Cx^3}
3. Practice Questions
Solve the following Bernoulli equations:
\frac{dy}{dx} + xy = xy^2\
\frac{dy}{dx} + xy = y^2 e^{\frac{x^2}{2}} \sin x\
Not sure about your answers? Don't worry, click the link below to check the video where the answers are explained.
Practice Questions Discussion
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