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#

Answer 1

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

We have:

# (x+2)siny \ dx + xcosy \ dy = 0 #

which, in standard form, is as follows:

# xcosy \ dy/dx + (x+2)siny = 0 #

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.

# \ \ \ \ \ cosy/siny \ dy/dx + (x+2)/x = 0 #
# :. int \ coty \ dy = - int \ (x+2)/x \ dx #

Moreover, we integrate to obtain:

# ln siny = -x - 2lnx + C#

By multiplying, we obtain:

# siny = e^(-x - 2lnx + C)# # \ \ \ \ \ \ \ = e^(C-x)e^(- 2lnx) # # \ \ \ \ \ \ \ = e^(C-x)e^(ln(1/x^2)) # # \ \ \ \ \ \ \ = e^(C-x)(1/x^2) # # \ \ \ \ \ \ \ = e^(C-x)/x^2 #

And lastly:

# y = arcsin(e^(C-x)/x^2) #
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

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:

  1. Identify the equation as non-exact.
  2. Determine the integrating factor.
  3. Multiply the entire equation by the integrating factor.
  4. 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.
  5. Check if the rewritten equation is exact.
  6. 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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7