Differential equation second order;]

[player1_1_1]

Homework Statement

I have differential equation;] this is equation from central force thing in Lagrange mechanics, you know, you know, its second order hahaha:D
$$y^3y^{\prime\prime}=ay+b$$

Homework Equations

I will using method of making a first degree of this equation

The Attempt at a Solution

I have
$$\frac{\mbox{d}y}{\mbox{d}x}=u(y)$$
$$\frac{\mbox{d}^2y}{\mbox{d}x^2}=\frac{\mbox{d}u}{\mbox{d}x}=\frac{\mbox{d}u}{\mbox{d}y}\frac{\mbox{d}y}{\mbox{d}x}=u\frac{\mbox{d}u}{\mbox{d}y}$$
and now I enter this new equation to before:
$$uy^3\frac{\mbox{d}u}{\mbox{d}y}=ay+b$$
$$u\frac{\mbox{d}u}{\mbox{d}y}=\frac{ay+b}{y^3}$$
$$u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C$$
$$u=\frac{\sqrt{Cy^2-2ay-4b}}{y}$$
$$\frac{y}{\sqrt{Cy^2-2ay-4b}}\frac{\mbox{d}y}{\mbox{d}x}=1$$
someone who is good in mathematic, please tell me if this calculations was good and what can I do now, I was thinking about do this with area functions, but not sure, please tell me simplest way to solve it;] thank you!

 

[Altabeh]

Homework Statement

I have differential equation;] this is equation from central force thing in Lagrange mechanics, you know, you know, its second order hahaha:D
$$y^3y^{\prime\prime}=ay+b$$

Homework Equations

I will using method of making a first degree of this equation

The Attempt at a Solution

I have
$$\frac{\mbox{d}y}{\mbox{d}x}=u(y)$$
$$\frac{\mbox{d}^2y}{\mbox{d}x^2}=\frac{\mbox{d}u}{\mbox{d}x}=\frac{\mbox{d}u}{\mbox{d}y}\frac{\mbox{d}y}{\mbox{d}x}=u\frac{\mbox{d}u}{\mbox{d}y}$$
and now I enter this new equation to before:
$$uy^3\frac{\mbox{d}u}{\mbox{d}y}=ay+b$$
$$u\frac{\mbox{d}u}{\mbox{d}y}=\frac{ay+b}{y^3}$$
$$u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C$$
$$u=\frac{\sqrt{Cy^2-2ay-4b}}{y}$$
$$\frac{y}{\sqrt{Cy^2-2ay-4b}}\frac{\mbox{d}y}{\mbox{d}x}=1$$
someone who is good in mathematic, please tell me if this calculations was good and what can I do now, I was thinking about do this with area functions, but not sure, please tell me simplest way to solve it;] thank you!

What is that -4 in $$u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C$$? It must be -1 since if you take the derivative of the second term wrt y, a 2y would appear in the numerator which cancels out a factor 1/2 coming from another side so the term b/y^3 is then retrieved. Now that dx should be brought to the other side of equation and then there you are left with a simple integration. Just note that since you are taking the root of $$u^2=-2\frac{a}{y}-4\frac{b}{y^2}+C$$, don't forget to have a sign $$\pm$$ accompanied with the radical.

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