# Non-exact 1st Order Differential Equation? Kindly solve this Non-exact 1st Order Differential Equation by integrating Factor Method. #(x+2)sinydx+x cosydy=0#

# y = arcsin(e^(C-x)/x^2) #

We have:

which, in standard form, is as follows:

Since this is a separable First Order differential equation, we can gather terms and separate the variables to obtain the solution without the need for an integrating factor.

Moreover, we integrate to obtain:

By multiplying, we obtain:

And lastly:

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To solve the non-exact first-order differential equation ( (x+2)\sin(y)dx + x\cos(y)dy = 0 ) using the integrating factor method, follow these steps:

- Identify the equation as non-exact.
- Determine the integrating factor.
- Multiply the entire equation by the integrating factor.
- Rewrite the equation in the form ( M(x,y)dx + N(x,y)dy = 0 ) where ( M ) and ( N ) are partial derivatives of the integrating factor multiplied by the original terms.
- Check if the rewritten equation is exact.
- If it's exact, solve it. If not, reconsider the steps.

Step 1: Identify the equation as non-exact.

The given equation ( (x+2)\sin(y)dx + x\cos(y)dy = 0 ) is non-exact.

Step 2: Determine the integrating factor ( \mu ).

The integrating factor ( \mu ) is given by: [ \mu = e^{\int P(x) , dx} ]

where ( P(x) ) is the coefficient of ( dx ).

In this case, ( P(x) = \sin(y) ), so: [ \mu = e^{\int \sin(y) , dx} = e^{-\cos(y)} ]

Step 3: Multiply the entire equation by the integrating factor.

Multiplying by ( \mu = e^{-\cos(y)} ): [ e^{-\cos(y)}(x+2)\sin(y)dx + e^{-\cos(y)}x\cos(y)dy = 0 ]

Step 4: Rewrite the equation in the form ( M(x,y)dx + N(x,y)dy = 0 ).

Let's denote ( M(x,y) = e^{-\cos(y)}(x+2)\sin(y) ) and ( N(x,y) = e^{-\cos(y)}x\cos(y) ).

So, the equation becomes: [ M(x,y)dx + N(x,y)dy = 0 ]

Step 5: Check if the rewritten equation is exact.

To check if the equation is exact, compute the partial derivatives ( \frac{\partial M}{\partial y} ) and ( \frac{\partial N}{\partial x} ) and verify if they are equal.

[ \frac{\partial M}{\partial y} = e^{-\cos(y)}\sin(y)(x+2) - e^{-\cos(y)}\sin(y)\sin(y)(x+2) ] [ \frac{\partial N}{\partial x} = e^{-\cos(y)}\cos(y) - e^{-\cos(y)}\cos(y)(x+2) ]

Step 6: Solve the equation.

Since the partial derivatives are not equal, the rewritten equation is still non-exact. Therefore, further methods like finding an integrating factor for exactness or using other techniques such as separation of variables may be necessary to solve this differential equation.

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