Why Can't I Solve y''=3/y²+5? Explained

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In summary: The solution of the first-order ODE is then given implicitly by x = \pm\int \frac{1}{\sqrt{2F(y)}} \,dy + C for some constant C. In summary, solving ODEs of the form y''(x) = f(y(x)) involves multiplying by y' and integrating to obtain a separable first-order ODE, which can be solved implicitly.
  • #1
Teacame
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If I have a DE like this:
[tex]y''=\frac{3}{y^{2}}+5[/tex]
Why can't I just integrate both sides to get:
[tex]y'=-\frac{3}{y^{1}}+5y[/tex]?
And then integrate again to solve for y?
 
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  • #2
Teacame said:
If I have a DE like this:
[tex]y''=\frac{3}{y^{2}}+5[/tex]
Why can't I just integrate both sides to get:
[tex]y'=-\frac{3}{y^{1}}+5y[/tex]?
And then integrate again to solve for y?

Because ##y''## means ##\frac{d^2y}{dt^2}##

So, you would need to integrate both sides wrt ##t##.

If you let ##\frac{d^2f}{dy^2} = \frac{3}{y^{2}}+5##

Where ##f## is a function of ##y##, then you can simply integrate both sides wrt ##y##.
 
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  • #3
PeroK said:
Because ##y''## means ##\frac{d^2y}{dt^2}##

So, you would need to integrate both sides wrt ##t##.

If you let ##\frac{d^2f}{dy^2} = \frac{3}{y^{2}}+5##

Where ##f## is a function of ##y##, then you can simply integrate both sides wrt ##y##.
Oh that was really dumb of me, didn't think about the notation enough. I'm not supposed to actually solve this, I was just wondering. It looks like the solution is extremely complicated so I probably can't anyway: http://www.wolframalpha.com/input/?i=y''=3/y^2+5

In fact, the solution for just y''=1/y looks complicated: http://www.wolframalpha.com/input/?i=y''=1/y

I haven't actually studied non-linear DEs yet.
 
  • #4
ODEs of the form [itex]y''(x) = f(y(x))[/itex] can in principle be solved by multiplying by [itex]y'[/itex] and integrating with respect to [itex]x[/itex] to obtain [tex]
\frac12(y')^2 = F(y)[/tex] where [itex]F(y) = \frac12 y'(0)^2 + \int_{y(0)}^{y} f(s)\,ds[/itex] is an antiderivative of [itex]f[/itex]. The resulting first-order ODE is separable: [tex]
y' = \pm\sqrt{2F(y)}
[/tex] where some care is needed in determining the sign.
 
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Related to Why Can't I Solve y''=3/y²+5? Explained

1. Why is it difficult to solve y''=3/y²+5?

This equation is difficult to solve because it is a second-order differential equation, meaning that it contains a second derivative of the dependent variable y. Solving second-order differential equations often requires advanced mathematical techniques and can be time-consuming.

2. Can't I just use basic algebra to solve y''=3/y²+5?

Unfortunately, basic algebra is not sufficient to solve this equation. As mentioned, it is a second-order differential equation and requires more advanced methods such as substitution or integration to solve.

3. What does the "y''" notation mean in the equation y''=3/y²+5?

The "y''" notation represents the second derivative of the dependent variable y. It indicates how the rate of change of y is changing. In simpler terms, it shows the acceleration of y.

4. Is there a specific method or formula to solve y''=3/y²+5?

Yes, there are various methods and formulas that can be used to solve second-order differential equations such as this one. Some common methods include substitution, separation of variables, and integrating factors. The most appropriate method to use may vary depending on the specific equation and its form.

5. Why is it important to solve y''=3/y²+5 in the first place?

Solving equations like this one can be important for various reasons. It may be necessary in order to understand and model real-world phenomena in fields such as physics, engineering, or economics. Additionally, solving this equation may lead to a better understanding of mathematical concepts and techniques, which can be useful in future problem-solving.

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