3y" + 6ỷ2=0
To solve the differential equation \(3y'' + 6y^2 = 0\), we first note that this is a second-order nonlinear ordinary differential equation.
Rearranging the Equation:
We can rewrite the equation as:
\[ y'' + 2y^2 = 0 \]
Assuming a Solution:
One method to solve such a differential equation is to make a substitution. However, this particular equation suggests we look for energy-like functions or conservation laws.
Multiply by \(y'\):
To facilitate integration, we multiply the entire equation by \(y'\):
\[ y'y'' + 2yy'^{2} = 0 \]
This form can be treated as a derivative:
\[ \frac{d}{dt} \left( \frac{1}{2}(y')^2 + y^2 \right) = 0 \]
From this, we can integrate to find:
\[ \frac{1}{2}(y')^2 + y^2 = C \]
where \(C\) is a constant of integration.
Rearranging the Conservation Law:
Rearranging gives:
\[ (y')^2 = 2C - 2y^2 \]
Taking the Square Root:
Now, taking the square root gives us:
\[ y' = \pm \sqrt{2C - 2y^2} \]
This is a separable equation which we can rearrange as:
\[ \frac{dy}{\sqrt{2C - 2y^2}} = \pm dt \]
Integrating:
For the left side, we recognize it as an integral that forms a trigonometric function:
\[ \int \frac{dy}{\sqrt{2C - 2y^2}} = \pm t + K \]
where \(K\) is a new constant. The integral evaluates to:
\[ \frac{1}{\sqrt{2}} \sin^{-1} \left(\frac{y}{\sqrt{C}}\right) \]
Thus, we have:
\[ \frac{1}{\sqrt{2}} \sin^{-1} \left(\frac{y}{\sqrt{C}}\right) = \pm t + K \]
Solving this leads to:
\[ y(t) = \sqrt{C} \sin(\sqrt{2}(\pm t + K)) \]
Conclusion:
The general solution to the differential equation \(3y'' + 6y^2 = 0\) is:
\[ y(t) = \sqrt{C} \sin(\sqrt{2}(\pm t + K)) \]
where \(C\) and \(K\) are constants determined by initial conditions.
If you have specific initial conditions, you can use them to find the particular solution.
