Wikipedia
solution of a differential equation : Many methods to compute numerical solutions of differential equations or study the properties of differential equations involve approximation of the ..... More from Wikipedia
solution of a differential equation : Any differential equation of order n can be written as a system of n first-order differential equations. ..... More from Wikipedia
General solution of a differential equation
If the solution of a differential equation of order n contains n arbitrary constants, then it is called the General solution of the differential equation..
Solution of a Differential Equation
Solution of a Differential Equation: The functional relation-ship between the independent variable and the dependent variable (such as y = f(x)) which satisfies the given differential equation is called the solution of the dif..
Power Series Solutions of Differential Equations - In this video, I show how to use power series to find a solution of a differential equation. This is a SIMPLE example and the final solution is very NICE compared to what would normally happen with a more complicated differential equation, so please be aware of that!
The video is a simple introduction to the area of "ordinary differential equations" (ODEs). We define what an ODE is and what `a solution' really means. The topic is motivated via simple examples and applications. In particular, ODEs are a powerful tool for modeling dynamical processes and for making precise predictions about future states of phenomena. Such ideas are seen in 1st year university mathematics courses.
Question : I need help finding the general solution for this differential equation:
y dx - 4(x + y^6) dy =0
Specifically, I'm having trouble identifying what P(x) and the integrating factor should be.
Thanks!
Answer : You'll have to re-write the equation to the standard first order nonlinear differential equation form to identify P(x), i.e. y' + P(x)y = g(x); where the integrating factor is u(x) = exp( P(x)dx); then multiplying both sides by u(x) to get u(x)y' + P(x)u(x)y = u(x)g(x) Notice that the LHS becomes an exact differential, d(u(x)y). Given: y dx - 4(x + y^6) dy =0 In this case, it would be easier to evaluate this using x as the dependent variable, i.e. we will solve for x(y) instead of the usual y(x) Divide both sides by dy; ydx/dy - 4x - 4y^6 = 0 yx' -4x = 4y^6 x' -4x/y = 4y^5 ----> (1) Clearly comparing to the standard form, P(y) = -4/y; therefore the integrating factor can be calculated as: u(y) = exp( P(y)dy) u(y) = exp( -4/y dy) = exp(ln(y^-4)) = 1/y^4 multiplying both sides of the differential eqn (1) by u(y) x'/y^4 - 4x/y^5 = 4y ---> here the LHS is now a total differential d(x/y^4)/dy d(x/y^4) = 4ydy integrating both sides, x/y^4 = 4(y^2)/2 + .... More from Yahoo Answers
Answer : You'll have to re-write the equation to the standard first order nonlinear differential equation form to identify P(x), i.e. y' + P(x)y = g(x); where the integrating factor is u(x) = exp( P(x)dx); then multiplying both sides by u(x) to get u(x)y' + P(x)u(x)y = u(x)g(x) Notice that the LHS becomes an exact differential, d(u(x)y). Given: y dx - 4(x + y^6) dy =0 In this case, it would be easier to evaluate this using x as the dependent variable, i.e. we will solve for x(y) instead of the usual y(x) Divide both sides by dy; ydx/dy - 4x - 4y^6 = 0 yx' -4x = 4y^6 x' -4x/y = 4y^5 ----> (1) Clearly comparing to the standard form, P(y) = -4/y; therefore the integrating factor can be calculated as: u(y) = exp( P(y)dy) u(y) = exp( -4/y dy) = exp(ln(y^-4)) = 1/y^4 multiplying both sides of the differential eqn (1) by u(y) x'/y^4 - 4x/y^5 = 4y ---> here the LHS is now a total differential d(x/y^4)/dy d(x/y^4) = 4ydy integrating both sides, x/y^4 = 4(y^2)/2 + .... More from Yahoo Answers
Question : What is the main contrast between a solution of an algebraic equation and a solution of a differential equation?
Answer : The main difference is that, in an algebraic equation, the variables represent values, whereas in a differential equation, the variables represent functions. So, an algebraic solution is somewhat one-dimensional: it represents a value, set of values, interval or intervals satisfying some constraints. (Of course, if the equation is underdetermined, the solution may be expressed in terms of other variables, but the solution set is still concerned with single-valued variables.) The solution of a differential equation is two- (or more-) dimensional--it represents a function or class of functions... More from Yahoo Answers
Answer : The main difference is that, in an algebraic equation, the variables represent values, whereas in a differential equation, the variables represent functions. So, an algebraic solution is somewhat one-dimensional: it represents a value, set of values, interval or intervals satisfying some constraints. (Of course, if the equation is underdetermined, the solution may be expressed in terms of other variables, but the solution set is still concerned with single-valued variables.) The solution of a differential equation is two- (or more-) dimensional--it represents a function or class of functions... More from Yahoo Answers
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