Solved on Feb 22, 2024

Identify the exact ordinary differential equation (ODE) from the given options.
(x+y)dx+(y+x)dy=0(x+y) dx + (y+x) dy = 0 (xy+y)dx+(xy+x)dy=0(xy+y) dx + (xy+x) dy = 0 (y+2)dx+(2x+1)dy=0(y+2) dx + (2x+1) dy = 0

STEP 1

Assumptions
1. We are given four different ordinary differential equations (ODEs).
2. We need to determine which of these is an exact differential equation (DE).
3. An exact DE has the form M(x,y)dx+N(x,y)dy=0M(x, y)dx + N(x, y)dy = 0, where MM and NN are continuous functions of xx and yy in a region RR, and there exists a function F(x,y)F(x, y) such that M=FxM = \frac{\partial F}{\partial x} and N=FyN = \frac{\partial F}{\partial y}.
4. For an exact DE, the condition My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} must hold in the region RR.

STEP 2

We will check each given ODE to see if it is an exact DE by verifying the condition My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.

STEP 3

Check if ODE (a) is exact:
For the ODE (x+y)dx+(y+x)dy=0(x+y)dx + (y+x)dy = 0, we identify M(x,y)=x+yM(x, y) = x + y and N(x,y)=y+xN(x, y) = y + x.

STEP 4

Calculate the partial derivative of MM with respect to yy:
My=y(x+y)=1\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x + y) = 1

STEP 5

Calculate the partial derivative of NN with respect to xx:
Nx=x(y+x)=1\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(y + x) = 1

STEP 6

Since My=Nx\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, ODE (a) is an exact DE.

STEP 7

We have found that ODE (a) is an exact DE, so there is no need to check the other ODEs. However, for completeness, we will verify whether the other ODEs are exact or not.

STEP 8

Check if ODE (b) is exact:
For the ODE (xy+y)dx+(xy+x)dy=0(xy + y)dx + (xy + x)dy = 0, we identify M(x,y)=xy+yM(x, y) = xy + y and N(x,y)=xy+xN(x, y) = xy + x.

STEP 9

Calculate the partial derivative of MM with respect to yy:
My=y(xy+y)=x+1\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(xy + y) = x + 1

STEP 10

Calculate the partial derivative of NN with respect to xx:
Nx=x(xy+x)=y+1\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xy + x) = y + 1

STEP 11

Since MyNx\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}, ODE (b) is not an exact DE.

STEP 12

Check if ODE (c) is exact:
For the ODE (y+2)dx+(2x+1)dy=0(y + 2)dx + (2x + 1)dy = 0, we identify M(x,y)=y+2M(x, y) = y + 2 and N(x,y)=2x+1N(x, y) = 2x + 1.

STEP 13

Calculate the partial derivative of MM with respect to yy:
My=y(y+2)=1\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y + 2) = 1

STEP 14

Calculate the partial derivative of NN with respect to xx:
Nx=x(2x+1)=2\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x + 1) = 2

STEP 15

Since MyNx\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}, ODE (c) is not an exact DE.

STEP 16

Since we have already found an exact DE in option (a), we conclude that option (d) "None" is not the correct answer.
The exact DE is option (a): (x+y)dx+(y+x)dy=0(x+y)dx + (y+x)dy = 0.

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