Riccati Equation/y' = (y over x) + x^3 y^2 - x^5

Theorem

The Riccati equation:

$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$

has the general solution:

$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$

Proof

It can be seen by inspection that:

$\map {y_1} x = x$

is a particular solution to $(1)$.

Thus from General Solution of Riccati Equation from Particular Solution:

$y = x + \map z x$

where:

$z' - \paren {\dfrac 1 x + 2 x^4} z = x^3 z^2$

From Bernoulli's Equation: $y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$:

$z = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$

Thus:

\(\ds y\)

\(=\)

\(\ds x + \frac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}\)

\(\ds \leadsto \ \ \)

\(\ds \paren {y - x} \paren {C \map \exp {-\dfrac {2 x^5} 5} - 1}\)

\(=\)

\(\ds 2 x\)

\(\ds \leadsto \ \ \)

\(\ds y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} + x\)

\(=\)

\(\ds 2 x\)

\(\ds \leadsto \ \ \)

\(\ds y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} - x\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \paren {y - x} C \map \exp {-\dfrac {2 x^5} 5}\)

\(=\)

\(\ds y + x\)

\(\ds \leadsto \ \ \)

\(\ds \frac {y - x} {y + x}\)

\(=\)

\(\ds C \map \exp {\dfrac {2 x^5} 5}\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $28$