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Solve the differential equation?

Solve the differential equation (2xy+y^3)dx+(x^2+3xy^2-2y)dy=0.

I got x^2*y+xy^3+yx^2+xy^3-y^2=C but that's not the answer. And is Differential Equations taught after Linear Algebra? If so, then is differential equations same as partial differential equations? Is partial differential equations a different course taught from differential equations? Which of these 2 comes first?

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Do you guys know how to input this equation on Ti-89 calculator to find the answer? I know how to do this problem but I want to know this technique to check other problems. Like you press F3, then C, deSolve...How do I put this equation?

I don't have T89 in my hands now, but Ias i rememeber, you have to go to MATH menu. It opens a llist of options. Scroll down the screen to get to to the line "differential equations". As I know, it opens the screen in which you can type your equation.

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2 Answers

You have to study linera algebra first, then course of differential equations, because some methods of differential eduaqtions are based on methods of linear algebra (use of determinants such as Wornscian, finding characteristic numbers of DEs, etc.). Equations with partial derivative are equations for function of multiple variable, and these equations are usually a part of mathematical physics.

If you notice that

                   y3dx + 3xy2dy = d(xy3),  2xydx + x2dy = d(x2y) ,    2ydy = d(y2)

then your differential equation can be written in the form

                                         d(xy3 + x2y) = d(y2)    or

                                          xy3 + x2y = y2 + C

 

 

You are close!

The form of the differential equation suggests a check for exactness.

(Df/Dx)dx + (Df/Dy)dy = 0  where 'D' is used as the symbol for partial differentiation.

Check if D(Df/Dx)/Dy = D(Df/Dy)/Dx.  In this case 2x + 3y^2 = 2x + 3y^2, so the differential equation is exact.

 

Df/Dx = 2xy + y^3 is to be integrated with respect to x to get f(x,y) = yx^2 + xy^3 + f(y)

Df/Dy = x^2  + 3xy^2-2y integrates to f(x,y) = yx^2 +xy^3 - y^2 + f(x)

 

Comparing the two solutions, f(y) = -y^2 and f(x) = 0.

 

The solution is C = yx^2 +xy^3 -y^2.  Although we work with partial derivatives, exact differential equations are usually taught in the differential equations.

 

In some college math departments, linear algebra is taught along with differential equations in the same class.  Some colleges require linear algebra or differential equations or both depending on a student's major.  The differential equations class usually follow a 2 or 3 semester sequence in calculus.  A basic linear algebra course could be taught that does not require prerequisites in calculus.  The differential equations class usually covers DE's with one independent variable (ordinary differential equations).  Maybe at the end of the course, some partial differential equations are introduced but are usually covered in an advanced courses.