Question:
Solve the separable differential equation Subject to the initial condition: , y = ?
(My work is attached)
Any help would be greatly appreciated!
Thanks in advance!
$\displaystyle 3x - 4y\sqrt{x^2 + 1}\,\frac{dy}{dx} = 0$
$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} - 4y\,\frac{dy}{dx} = 0$
$\displaystyle \frac{3x}{\sqrt{x^2 + 1}} = 4y\,\frac{dy}{dx}$
$\displaystyle \int{\frac{3x}{\sqrt{x^2 + 1}}\,dx} = \int{4y\,\frac{dy}{dx}\,dx}$
$\displaystyle \int{3x(x^2 + 1)^{-\frac{1}{2}}\,dx} = \int{4y\,dy}$.
You should be able to evaluate both integrals now. To do the left hand side, you need to use the substitution $\displaystyle u = x^2 + 1$ so that $\displaystyle \frac{du}{dx} = 2x$.
1. Don't be so rude - most people on this forum do not open attachments.
2. When you have got it to the step:
$\displaystyle \frac{3}{2}\int{u^{-\frac{1}{2}}\,du} = \int{4y\,dy}$
$\displaystyle 3u^{\frac{1}{2}} + C_1 = 2y^2 + C_2$
$\displaystyle 3\sqrt{x^2 + 1} + C_1 - C_2 = 2y^2$
$\displaystyle \frac{3}{2}\sqrt{x^2 + 1} + C = y^2$, where $\displaystyle C = \frac{1}{2}(C_1 - C_2)$
$\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} + C}$.
If $\displaystyle y(0) = 1$
$\displaystyle 1 = \pm \sqrt{\frac{3}{2}\sqrt{0^2 + 1} + C}$
$\displaystyle 1 = \frac{3}{2}\sqrt{1} + C$
$\displaystyle 1 = \frac{3}{2} + C$
$\displaystyle C = -\frac{1}{2}$.
So $\displaystyle y = \pm \sqrt{\frac{3}{2}\sqrt{x^2 + 1} - \frac{1}{2}}$
$\displaystyle y = \pm \sqrt{\frac{3\sqrt{x^2 + 1} - 1}{2}}$
$\displaystyle y = \pm \frac{\sqrt{3\sqrt{x^2 + 1} - 1}}{\sqrt{2}}$
$\displaystyle y = \pm \frac{\sqrt{6\sqrt{x^2 + 1} - 2}}{2}$.