1.
( 2 x + 4 y − 5 ) d x + ( 6 y + 4 x − 1 ) d y = 0 (2x + 4y − 5)dx + (6y + 4x − 1)dy = 0 ( 2 x + 4 y − 5 ) d x + ( 6 y + 4 x − 1 ) d y = 0
∂ Q ∂ x = 4 , ∂ P ∂ y = 4 \dfrac{\partial Q}{\partial x}=4, \dfrac{\partial P}{\partial y}=4 ∂ x ∂ Q = 4 , ∂ y ∂ P = 4 The given equation is exact because the partial derivatives are the same:
∂ Q ∂ x = 4 = ∂ P ∂ y \dfrac{\partial Q}{\partial x}=4=\dfrac{\partial P}{\partial y} ∂ x ∂ Q = 4 = ∂ y ∂ P We have the following system of differential equations to find the function u ( x , y ) u(x, y) u ( x , y )
{ ∂ u ∂ x = 2 x + 4 y − 5 ∂ u ∂ y = 6 y + 4 x − 1 \begin{cases}
\dfrac{\partial u}{\partial x}=2x + 4y − 5 \\
\\
\dfrac{\partial u}{\partial y}=6y + 4x − 1
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ u = 2 x + 4 y − 5 ∂ y ∂ u = 6 y + 4 x − 1
u ( x , y ) = ∫ ( 2 x + 4 y − 5 ) d x = x 2 + 4 x y − 5 x + φ ( y ) u(x, y)=\int (2x + 4y − 5)dx=x^2+4xy-5x+\varphi (y) u ( x , y ) = ∫ ( 2 x + 4 y − 5 ) d x = x 2 + 4 x y − 5 x + φ ( y )
∂ u ∂ y = 4 x + φ ′ ( y ) = 6 y + 4 x − 1 \dfrac{\partial u}{\partial y}=4x+\varphi' (y)=6y + 4x − 1 ∂ y ∂ u = 4 x + φ ′ ( y ) = 6 y + 4 x − 1
φ ′ ( y ) = 6 y − 1 \varphi' (y)=6y − 1 φ ′ ( y ) = 6 y − 1
The general solution of the exact differential equation is given by
x 2 + 4 x y − 5 x + 3 y 2 − y = C x^2+4xy-5x+3y^2-y=C x 2 + 4 x y − 5 x + 3 y 2 − y = C where C C C
2.
( y + 2 x y 3 ) d x + ( 1 + 3 x 2 y 2 + x ) d y = 0 (y + 2xy^3)dx + (1+ 3x^2y^2 + x)dy = 0 ( y + 2 x y 3 ) d x + ( 1 + 3 x 2 y 2 + x ) d y = 0
∂ Q ∂ x = 6 x y 2 + 1 , ∂ P ∂ y = 1 + 6 x y 2 \dfrac{\partial Q}{\partial x}=6xy^2+1, \dfrac{\partial P}{\partial y}=1+6xy^2 ∂ x ∂ Q = 6 x y 2 + 1 , ∂ y ∂ P = 1 + 6 x y 2 The given equation is exact because the partial derivatives are the same:
∂ Q ∂ x = 6 x y 2 + 1 = ∂ P ∂ y \dfrac{\partial Q}{\partial x}=6xy^2+1=\dfrac{\partial P}{\partial y} ∂ x ∂ Q = 6 x y 2 + 1 = ∂ y ∂ P We have the following system of differential equations to find the function u ( x , y ) u(x, y) u ( x , y )
{ ∂ u ∂ x = y + 2 x y 3 ∂ u ∂ y = 1 + 3 x 2 y 2 + x \begin{cases}
\dfrac{\partial u}{\partial x}=y + 2xy^3 \\
\\
\dfrac{\partial u}{\partial y}=1+ 3x^2y^2 + x
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ u = y + 2 x y 3 ∂ y ∂ u = 1 + 3 x 2 y 2 + x
u ( x , y ) = ∫ ( y + 2 x y 3 ) d x = x y + x 2 y 3 + φ ( y ) u(x, y)=\int (y + 2xy^3)dx=xy+x^2y^3+\varphi (y) u ( x , y ) = ∫ ( y + 2 x y 3 ) d x = x y + x 2 y 3 + φ ( y )
∂ u ∂ y = x + 3 x 2 y 2 + φ ′ ( y ) = 1 + 3 x 2 y 2 + x \dfrac{\partial u}{\partial y}=x+3x^2y^2+\varphi' (y)=1+ 3x^2y^2 + x ∂ y ∂ u = x + 3 x 2 y 2 + φ ′ ( y ) = 1 + 3 x 2 y 2 + x
φ ′ ( y ) = 1 \varphi' (y)= 1 φ ′ ( y ) = 1
The general solution of the exact differential equation is given by
x y + x 2 y 3 + y = C xy+x^2y^3+y=C x y + x 2 y 3 + y = C where C C C
3.
3 x 2 y d x + ( x 3 + 2 y 4 ) d y = 0 3x^2ydx + (x^3+2y^4)dy = 0 3 x 2 y d x + ( x 3 + 2 y 4 ) d y = 0
∂ Q ∂ x = 3 x 2 , ∂ P ∂ y = 3 x 2 \dfrac{\partial Q}{\partial x}=3x^2, \dfrac{\partial P}{\partial y}=3x^2 ∂ x ∂ Q = 3 x 2 , ∂ y ∂ P = 3 x 2 The given equation is exact because the partial derivatives are the same:
∂ Q ∂ x = 3 x 2 = ∂ P ∂ y \dfrac{\partial Q}{\partial x}=3x^2=\dfrac{\partial P}{\partial y} ∂ x ∂ Q = 3 x 2 = ∂ y ∂ P We have the following system of differential equations to find the function u ( x , y ) u(x, y) u ( x , y )
{ ∂ u ∂ x = 3 x 2 y ∂ u ∂ y = x 3 + 2 y 4 \begin{cases}
\dfrac{\partial u}{\partial x}=3x^2y \\
\\
\dfrac{\partial u}{\partial y}=x^3+2y^4
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ u = 3 x 2 y ∂ y ∂ u = x 3 + 2 y 4
u ( x , y ) = ∫ 3 x 2 y d x = x 3 y + φ ( y ) u(x, y)=\int 3x^2ydx=x^3y+\varphi (y) u ( x , y ) = ∫ 3 x 2 y d x = x 3 y + φ ( y )
∂ u ∂ y = x 3 + φ ′ ( y ) = x 3 + 2 y 4 \dfrac{\partial u}{\partial y}=x^3+\varphi' (y)=x^3+2y^4 ∂ y ∂ u = x 3 + φ ′ ( y ) = x 3 + 2 y 4
φ ′ ( y ) = 2 y 4 \varphi' (y)=2y^4 φ ′ ( y ) = 2 y 4
The general solution of the exact differential equation is given by
x 3 y + 2 5 y 5 = C x^3y+\dfrac{2}{5}y^5=C x 3 y + 5 2 y 5 = C where C C C
4.
( 1 y ) d x − ( x y 2 ) d y = 0 (\dfrac{1}{y})dx - (\dfrac{x}{y^2})dy = 0 ( y 1 ) d x − ( y 2 x ) d y = 0
∂ Q ∂ x = − 1 y 2 , ∂ P ∂ y = − 1 y 2 \dfrac{\partial Q}{\partial x}=-\dfrac{1}{y^2}, \dfrac{\partial P}{\partial y}=-\dfrac{1}{y^2} ∂ x ∂ Q = − y 2 1 , ∂ y ∂ P = − y 2 1 The given equation is exact because the partial derivatives are the same:
∂ Q ∂ x = − 1 y 2 = ∂ P ∂ y \dfrac{\partial Q}{\partial x}=-\dfrac{1}{y^2}=\dfrac{\partial P}{\partial y} ∂ x ∂ Q = − y 2 1 = ∂ y ∂ P We have the following system of differential equations to find the function u ( x , y ) u(x, y) u ( x , y )
{ ∂ u ∂ x = 1 y ∂ u ∂ y = − x y 2 \begin{cases}
\dfrac{\partial u}{\partial x}=\dfrac{1}{y} \\
\\
\dfrac{\partial u}{\partial y}=-\dfrac{x}{y^2}
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ u = y 1 ∂ y ∂ u = − y 2 x
u ( x , y ) = ∫ ( 1 y ) d x = x y + φ ( y ) u(x, y)=\int (\dfrac{1}{y})dx=\dfrac{x}{y}+\varphi (y) u ( x , y ) = ∫ ( y 1 ) d x = y x + φ ( y )
∂ u ∂ y = − x y 2 + φ ′ ( y ) = − x y 2 \dfrac{\partial u}{\partial y}=-\dfrac{x}{y^2}+\varphi' (y)=-\dfrac{x}{y^2} ∂ y ∂ u = − y 2 x + φ ′ ( y ) = − y 2 x
φ ′ ( y ) = 0 \varphi' (y)=0 φ ′ ( y ) = 0
The general solution of the exact differential equation is given by
x y = C \dfrac{x}{y}=C y x = C where C C C
5.
( 4 x 3 − y 3 ) d x − ( 3 x y 2 ) d y = 0 (4x^3-y^3)dx - (3xy^2)dy = 0 ( 4 x 3 − y 3 ) d x − ( 3 x y 2 ) d y = 0
∂ Q ∂ x = − 3 y 2 , ∂ P ∂ y = − 3 y 2 \dfrac{\partial Q}{\partial x}=-3y^2, \dfrac{\partial P}{\partial y}=-3y^2 ∂ x ∂ Q = − 3 y 2 , ∂ y ∂ P = − 3 y 2 The given equation is exact because the partial derivatives are the same:
∂ Q ∂ x = − 3 y 2 = ∂ P ∂ y \dfrac{\partial Q}{\partial x}=-3y^2=\dfrac{\partial P}{\partial y} ∂ x ∂ Q = − 3 y 2 = ∂ y ∂ P We have the following system of differential equations to find the function u ( x , y ) u(x, y) u ( x , y )
{ ∂ u ∂ x = 4 x 3 − y 3 ∂ u ∂ y = − 3 x y 2 \begin{cases}
\dfrac{\partial u}{\partial x}=4x^3-y^3 \\
\\
\dfrac{\partial u}{\partial y}=-3xy^2
\end{cases} ⎩ ⎨ ⎧ ∂ x ∂ u = 4 x 3 − y 3 ∂ y ∂ u = − 3 x y 2
u ( x , y ) = ∫ ( 4 x 3 − y 3 ) d x = x 4 − x y 3 + φ ( y ) u(x, y)=\int (4x^3-y^3)dx=x^4-xy^3+\varphi (y) u ( x , y ) = ∫ ( 4 x 3 − y 3 ) d x = x 4 − x y 3 + φ ( y )
∂ u ∂ y = − 3 x y 2 + φ ′ ( y ) = − 3 x y 2 \dfrac{\partial u}{\partial y}=-3xy^2+\varphi' (y)=-3xy^2 ∂ y ∂ u = − 3 x y 2 + φ ′ ( y ) = − 3 x y 2
φ ′ ( y ) = 0 \varphi' (y)=0 φ ′ ( y ) = 0
The general solution of the exact differential equation is given by
x 4 − x y 3 = C x^4-xy^3=C x 4 − x y 3 = C where C C C
#340153 on Dec 2023
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