Lets rewrite the equation as below:
( x y 2 − x 2 ) d x + ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d y = 0 p ( x , y ) d x + q ( x , y ) d y = 0 { p ( x , y ) = x y 2 − x 2 q ( x , y ) = 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 \left(xy^2-x^2\right)dx+\left(3x^2y^2+x^2y-2x^3+y^2\right)dy=0\\[0.3cm]
p(x,y)dx+q(x,y)dy=0\\[0.3cm]
\left\{\begin{array}{l}
p(x,y)=xy^2-x^2\\[0.3cm]
q(x,y)=3x^2y^2+x^2y-2x^3+y^2
\end{array}\right. ( x y 2 − x 2 ) d x + ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d y = 0 p ( x , y ) d x + q ( x , y ) d y = 0 { p ( x , y ) = x y 2 − x 2 q ( x , y ) = 3 x 2 y 2 + x 2 y − 2 x 3 + y 2
The equation is called exact differential equation if it satisfied:
∂ p ( x , y ) ∂ y = ∂ q ( x , y ) ∂ y \frac{\partial p(x,y)}{\partial y}=\frac{\partial q(x,y)}{\partial y} ∂ y ∂ p ( x , y ) = ∂ y ∂ q ( x , y )
In our case,
{ ∂ p ∂ y = 2 x y ∂ q ∂ x = 6 x y 2 + 2 x y − 6 x 2 ⟶ ∂ p ∂ y ≠ ∂ q ∂ x \left\{\begin{array}{l}
\displaystyle\frac{\partial p}{\partial y}=2xy\\[0.3cm]
\displaystyle\frac{\partial q}{\partial x}=6xy^2+2xy-6x^2
\end{array}\right.\longrightarrow\\[0.3cm]
\frac{\partial p}{\partial y}\neq\frac{\partial q}{\partial x} ⎩ ⎨ ⎧ ∂ y ∂ p = 2 x y ∂ x ∂ q = 6 x y 2 + 2 x y − 6 x 2 ⟶ ∂ y ∂ p = ∂ x ∂ q
Since the relation is not equal, this differential equation is not exact.
Next step is find a function u ( x , y ) u(x,y) u ( x , y )
u ( x , y ) ( p ( x , y ) d x + q ( x , y ) d y ) = 0 u(x,y)(p(x,y)dx+q(x,y)dy)=0 u ( x , y ) ( p ( x , y ) d x + q ( x , y ) d y ) = 0
u ( x , y ) u(x,y) u ( x , y ) x x x y y y
u ( x , y ) u(x,y) u ( x , y )
u ( x , y ) ≡ u ( x ) = ∂ p ∂ y − ∂ q ∂ x q ( x , y ) u(x,y)\equiv u(x)=\frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)} u ( x , y ) ≡ u ( x ) = q ( x , y ) ∂ y ∂ p − ∂ x ∂ q
only have x x x y y y
u ( x , y ) ≡ u ( y ) = ∂ q ∂ x − ∂ p ∂ y p ( x , y ) u(x,y)\equiv u(y)=\frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)} u ( x , y ) ≡ u ( y ) = p ( x , y ) ∂ x ∂ q − ∂ y ∂ p
only have y y y
In our case,
∂ p ∂ y − ∂ q ∂ x q ( x , y ) = 2 x y − ( 6 x y 2 + 2 x y − 6 x 2 ) 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ∂ p ∂ y − ∂ q ∂ x q ( x , y ) = 6 x ( x − y 2 ) 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 \frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)}=\frac{2xy-\left(6xy^2+2xy-6x^2\right)}{3x^2y^2+x^2y-2x^3+y^2}\\[0.3cm]
\frac{\displaystyle\frac{\partial p}{\partial y}-\displaystyle\frac{\partial q}{\partial x}}{q(x,y)}=\frac{6x\left(x-y^2\right)}{3x^2y^2+x^2y-2x^3+y^2}\\[0.3cm] q ( x , y ) ∂ y ∂ p − ∂ x ∂ q = 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 2 x y − ( 6 x y 2 + 2 x y − 6 x 2 ) q ( x , y ) ∂ y ∂ p − ∂ x ∂ q = 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 6 x ( x − y 2 )
contain both x x x y y y
∂ q ∂ x − ∂ p ∂ y p ( x , y ) = ( 6 x y 2 + 2 x y − 6 x 2 ) − 2 x y x y 2 − x 2 ∂ q ∂ x − ∂ p ∂ y p ( x , y ) = 6 x ( y 2 − x ) x ( y 2 − x ) = 6 \frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)}=\frac{\left(6xy^2+2xy-6x^2\right)-2xy}{xy^2-x^2}\\[0.3cm]
\frac{\displaystyle\frac{\partial q}{\partial x}-\displaystyle\frac{\partial p}{\partial y}}{p(x,y)}=\frac{6x\left(y^2-x\right)}{x(y^2-x)}=6\\[0.3cm] p ( x , y ) ∂ x ∂ q − ∂ y ∂ p = x y 2 − x 2 ( 6 x y 2 + 2 x y − 6 x 2 ) − 2 x y p ( x , y ) ∂ x ∂ q − ∂ y ∂ p = x ( y 2 − x ) 6 x ( y 2 − x ) = 6 contain only y y y
So u ( x , y ) u(x,y) u ( x , y ) y y y
u ( y ) = e ∫ 6 d y = e 6 y u(y)=e^{\int 6dy}=e^{6y} u ( y ) = e ∫ 6 d y = e 6 y
Back to our initial equation:
u ( y ) ( p ( x , y ) d x + q ( x , y ) d y ) = 0 e 6 y ( x y 2 − x 2 ) d x + e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d y = 0 u(y)(p(x,y)dx+q(x,y)dy)=0\\[0.3cm]
e^{6y}\left(xy^2-x^2\right)dx+e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right)dy=0 u ( y ) ( p ( x , y ) d x + q ( x , y ) d y ) = 0 e 6 y ( x y 2 − x 2 ) d x + e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d y = 0
Now find a function f ( x , y ) f(x,y) f ( x , y )
{ ∂ f ∂ x = e 6 y ( x y 2 − x 2 ) ∂ f ∂ y = e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) ∂ f ∂ x = e 6 y ( x y 2 − x 2 ) ⟶ f ( x , y ) = ∫ ( e 6 y ( x y 2 − x 2 ) ) d x + g ( y ) ⟶ f ( x , y ) = e 6 y ( x 2 y 2 2 − x 3 3 ) + g ( y ) ∂ f ∂ y = ∂ ∂ y ( e 6 y ( x 2 y 2 2 − x 3 3 ) + g ( y ) ) = = 6 ⋅ e 6 y ( x 2 y 2 2 − x 3 3 ) + e 6 y ( x 2 y ) + d g d y = = e 6 y ( 3 x 2 y 2 − 2 x 3 + x 2 y ) + d g d y \left\{\begin{array}{l}
\displaystyle\frac{\partial f}{\partial x}=e^{6y}\left(xy^2-x^2\right)\\[0.3cm]
\displaystyle\frac{\partial f}{\partial y}=e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right)
\end{array}\right.\\[0.3cm]
\frac{\partial f}{\partial x}=e^{6y}\left(xy^2-x^2\right)\longrightarrow\\[0.3cm]
f(x,y)=\int\left(e^{6y}\left(xy^2-x^2\right)\right)dx+g(y)\longrightarrow\\[0.3cm]
\boxed{f(x,y)=e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+g(y)}\\[0.3cm]
\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\left(e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+g(y)\right)=\\[0.3cm]
=6\cdot e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+e^{6y}\left(x^2y\right)+\frac{dg}{dy}=\\[0.3cm]
=e^{6y}\left(3x^2y^2-2x^3+x^2y\right)+\frac{dg}{dy} ⎩ ⎨ ⎧ ∂ x ∂ f = e 6 y ( x y 2 − x 2 ) ∂ y ∂ f = e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) ∂ x ∂ f = e 6 y ( x y 2 − x 2 ) ⟶ f ( x , y ) = ∫ ( e 6 y ( x y 2 − x 2 ) ) d x + g ( y ) ⟶ f ( x , y ) = e 6 y ( 2 x 2 y 2 − 3 x 3 ) + g ( y ) ∂ y ∂ f = ∂ y ∂ ( e 6 y ( 2 x 2 y 2 − 3 x 3 ) + g ( y ) ) = = 6 ⋅ e 6 y ( 2 x 2 y 2 − 3 x 3 ) + e 6 y ( x 2 y ) + d y d g = = e 6 y ( 3 x 2 y 2 − 2 x 3 + x 2 y ) + d y d g Then,
e 6 y ( 3 x 2 y 2 − 2 x 3 + x 2 y ) + d g d y = e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d g d y = e 6 y ⋅ y 2 ⟶ g ( y ) = ∫ ( e 6 y y 2 ) d y ( integration by parts ) : g ( y ) = ∫ ( e 6 y y 2 ) d y = y 2 e 6 y 6 − ∫ 2 e 6 y 6 d y = = y 2 e 6 y 6 − 1 3 ⋅ ( y e 6 y 6 − ∫ e 6 y 6 ) = = y 2 e 6 y 6 − y e 6 y 18 − e 6 y 108 ≡ e 6 y ( 18 y 2 − 6 y + 1 ) 108 e^{6y}\left(3x^2y^2-2x^3+x^2y\right)+\frac{dg}{dy}=e^{6y}\left(3x^2y^2+x^2y-2x^3+y^2\right)\\[0.3cm]
\frac{dg}{dy}=e^{6y}\cdot y^2\longrightarrow g(y)=\int\left(e^{6y}y^2\right)dy\\[0.3cm]
\left(\text{integration by parts}\right) : \\[0.3cm]
g(y)=\int\left(e^{6y}y^2\right)dy=\frac{y^2e^{6y}}{6}-\int\frac{2e^{6y}}{6}dy=\\[0.3cm]
=\frac{y^2e^{6y}}{6}-\frac{1}{3}\cdot\left(\frac{ye^{6y}}{6}-\int\frac{e^{6y}}{6}\right)=\\[0.3cm]
=\frac{y^2e^{6y}}{6}-\frac{ye^{6y}}{18}-\frac{e^{6y}}{108}\equiv\frac{e^{6y}\left(18y^2-6y+1\right)}{108} e 6 y ( 3 x 2 y 2 − 2 x 3 + x 2 y ) + d y d g = e 6 y ( 3 x 2 y 2 + x 2 y − 2 x 3 + y 2 ) d y d g = e 6 y ⋅ y 2 ⟶ g ( y ) = ∫ ( e 6 y y 2 ) d y ( integration by parts ) : g ( y ) = ∫ ( e 6 y y 2 ) d y = 6 y 2 e 6 y − ∫ 6 2 e 6 y d y = = 6 y 2 e 6 y − 3 1 ⋅ ( 6 y e 6 y − ∫ 6 e 6 y ) = = 6 y 2 e 6 y − 18 y e 6 y − 108 e 6 y ≡ 108 e 6 y ( 18 y 2 − 6 y + 1 )
Conclusion,
f ( x , y ) = f ( x , y ) = e 6 y ( x 2 y 2 2 − x 3 3 ) + e 6 y ( 18 y 2 − 6 y + 1 ) 108 \boxed{f(x,y)=f(x,y)=e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+\frac{e^{6y}\left(18y^2-6y+1\right)}{108}} f ( x , y ) = f ( x , y ) = e 6 y ( 2 x 2 y 2 − 3 x 3 ) + 108 e 6 y ( 18 y 2 − 6 y + 1 )
And finally our equation can be rewriten as:
∂ f ( x , y ) ∂ x ⋅ d x + ∂ f ( x , y ) ∂ y ⋅ d y = d f ( x , y ) = 0 f ( x , y ) = C e 6 y ( x 2 y 2 2 − x 3 3 ) + e 6 y ( 18 y 2 − 6 y + 1 ) 108 = C \frac{\partial f(x,y)}{\partial x}\cdot dx+\frac{\partial f(x,y)}{\partial y}\cdot dy=df(x,y)=0\\[0.3cm]
f(x,y)=C\\[0.3cm]
\boxed{e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}\right)+\frac{e^{6y}\left(18y^2-6y+1\right)}{108}=C} ∂ x ∂ f ( x , y ) ⋅ d x + ∂ y ∂ f ( x , y ) ⋅ d y = df ( x , y ) = 0 f ( x , y ) = C e 6 y ( 2 x 2 y 2 − 3 x 3 ) + 108 e 6 y ( 18 y 2 − 6 y + 1 ) = C
ANSWER
General solution
e 6 y ( x 2 y 2 2 − x 3 3 + 18 y 2 − 6 y + 1 108 ) = C e^{6y}\left(\frac{x^2y^2}{2}-\frac{x^3}{3}+\frac{18y^2-6y+1}{108}\right)=C e 6 y ( 2 x 2 y 2 − 3 x 3 + 108 18 y 2 − 6 y + 1 ) = C
#339914 on Dec 2023
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