Bernoulli's Equation/x y' + y = x^4 y^3
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Theorem
The first order ODE:
- $(1): \quad x y' + y = x^4 y^3$
has the general solution:
- $\dfrac 1 {y^2} = - x^4 + C x^2$
Proof
Let $(1)$ be rearranged as:
- $(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$
It can be seen that $(2)$ is in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
where:
- $\map P x = \dfrac 1 x$
- $\map Q x = x^3$
- $n = 3$
and so is an example of Bernoulli's equation.
By Solution to Bernoulli's Equation it has the general solution:
- $(3): \quad \ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
where:
- $\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
Thus $\map \mu x$ is evaluated:
\(\ds \paren {1 - n} \int \map P x \rd x\) | \(=\) | \(\ds \paren {1 - 3} \int \dfrac 1 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \ln x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {\frac 1 {x^2} }\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \mu x\) | \(=\) | \(\ds e^{\map \ln {\frac 1 {x^2} } }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {x^2}\) |
and so substituting into $(3)$:
\(\ds \frac 1 {x^2} \frac 1 {y^2}\) | \(=\) | \(\ds \paren {1 - 3} \int x^3 \frac 1 {x^2} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -2 \int x \rd x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {x^2 y^2}\) | \(=\) | \(\ds -2 \dfrac {x^2} 2 + C\) |
Hence the general solution to $(1)$ is:
- $\dfrac 1 {y^2} = - x^4 + C x^2$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.10$: Problem $3 \ \text{(a)}$