DE Identifying/Solving Homo/Bernoullis/Exact/etc

[netapparition]

Can anyone confirm that I have identified the proper types of Differential Equations and if anyone has the time to write out a decent notation of how to solve the following sampel questions I would appreciate it. We are using the Elementary Differential Equations Eigth Edition book by Boyle and DiPrima which has inconsistant notation.

a. (dy/dx)+y=y^2(cos(x)-sin(x)) => Bernoulli equation

b. (((sin(y)*(e^x))+e^(-y))dx-((xe^-y)-((e^x)(cosy))dy=0 => Exact Equation which needs an integrating factor.

c. (x^2+y^2+x)dx+xydy=0 => Homogeneous equation

d. (y^2-xy)dx+x^2dy=0 which has an integrating factor of the form u(x,y)=(x^m)(y^n). => I am unsure what type of equation this is.

If anyone can help please advise. Post or reply to netapparition@yahoo.com

 

[M98Ranger]

a. To solve this Bernoulli equation, first divide both sides by y^2:

(dy/dx)y^-2 + y^-1 = cos(x) - sin(x)

Then, let u = y^-1 and du/dx = -y^-2(dy/dx). Substituting this into the equation, we get:

-du/dx + u = cos(x) - sin(x)

This equation is now in the form of a linear differential equation, which can be solved by finding the integrating factor and using the method of integrating factors. The integrating factor is e^x, so multiply both sides by e^x:

e^x(-du/dx + u) = e^x(cos(x) - sin(x))

Then, use the product rule to simplify the left side:

d/dx(e^xu) = e^x(cos(x) - sin(x))

Integrating both sides with respect to x, we get:

e^xu = ∫e^x(cos(x) - sin(x))dx

Using integration by parts on the right side, we get:

e^xu = e^x(sin(x) + cos(x)) + C

Substituting back in u = y^-1, we get:

y^-1 = e^-x(sin(x) + cos(x)) + C

Solving for y, we get the general solution:

y = (e^-x(sin(x) + cos(x)) + C)^-1

b. To solve this exact equation, first check if the equation satisfies the exact equation condition:

Mx + Ny + ∂M/∂y = ∂N/∂x

Where M and N are the coefficients of dx and dy, respectively. In this case, M = (sin(y)e^x + e^-y) and N = -(xe^-y + e^xcos(y)). Taking partial derivatives, we get:

∂M/∂y = cos(y)e^x - e^-y

∂N/∂x = e^-y - e^xsin(y)

Since these are not equal, we need to find an integrating factor to make the equation exact. The integrating factor can be found by dividing the two partial derivatives and integrating with respect to either x or y. In this case, we will integrate with respect to x:

∫(∂M/∂y)/∂N

 

[M98Ranger]

Hello,

I am happy to assist with identifying and solving the types of differential equations mentioned in your post. After reviewing the equations provided, I can confirm that your identifications are correct. I have also included the notation and steps for solving each equation below.

a. (dy/dx)+y=y^2(cos(x)-sin(x)) => Bernoulli equation

To solve this equation, we can use the substitution u = y^(-1). This will transform the equation into a linear form, which can be easily solved using standard methods. The steps are as follows:

1. Substitute u = y^(-1) into the equation. This gives us: (dy/dx) + uy = cos(x) - sin(x)

2. Multiply both sides by u to get rid of the fraction. This gives us: (u(dy/dx)) + u^2y = u(cos(x) - sin(x))

3. Use the product rule to expand the first term on the left side. This gives us: (dy/dx) + u(dy/dx) + u^2y = u(cos(x) - sin(x))

4. Group the terms with dy/dx together and factor out u. This gives us: (dy/dx)(1 + u) + u^2y = u(cos(x) - sin(x))

5. Use the substitution u = y^(-1) to replace the u^2y term. This gives us: (dy/dx)(1 + u) + u(cos(x) - sin(x)) = u(cos(x) - sin(x))

6. Subtract u(cos(x) - sin(x)) from both sides to isolate the dy/dx term. This gives us: (dy/dx)(1 + u) = 0

7. Divide both sides by (1 + u) to solve for dy/dx. This gives us: dy/dx = 0

8. Integrate both sides with respect to x. This gives us: y = C, where C is a constant.

b. (((sin(y)*(e^x))+e^(-y))dx-((xe^-y)-((e^x)(cosy))dy=0 => Exact Equation which needs an integrating factor.

To solve this equation, we first need to check if it is exact. This can be done by checking if the partial derivatives of the terms with respect to x and y are