Question #fd5f0

Jun 7, 2017

I'll do the first one to show the general method of solving differential equations of this form.

These are all separable differential equations, meaning that all terms with $x$, including $\text{d} x$, can be moved to one side of the equation, while all terms including $y$ can be moved to the other.

From there, both sides can be integrated independently, and the constant of integration found using the initial condition.

Depending on context, it may or may not be possible to solve for $y$ explicitly.

In the first example

$\left(\text{d"y)/("d} x\right) = \frac{x}{y} ^ 2$

we can cross multiply to isolate $x$ terms and $y$ terms as:

${y}^{2} \textcolor{w h i t e}{.} \text{d"y=xcolor(white)."d} x$

Integrate both sides:

$\int {y}^{2} \textcolor{w h i t e}{.} \text{d"y=intxcolor(white)."d} x$

$\frac{1}{3} {y}^{3} = \frac{1}{2} {x}^{2} + C$

Note that $C$, an arbitrary constant of integration, has been added to only the right-hand side of the equation. This is arbitrary—we just as easily could have written $\left(1 / 3\right) {y}^{3} + {C}_{1} = \left(1 / 2\right) {x}^{2} + {C}_{2}$ or $\left(1 / 3\right) {y}^{3} + {C}_{3} = \left(1 / 2\right) {x}^{2}$. All are analogous forms.

Proceeding with the more standard form, which is to include the constant with the $x$ terms, we can solve for $C$ using the initial condition—that when $x = 1$, $y = 1$ as well.

$\frac{1}{3} {\left(1\right)}^{3} = \frac{1}{2} {\left(1\right)}^{2} + C$

$C = - \frac{1}{6}$

So:

$\frac{1}{3} {y}^{3} = \frac{1}{2} {x}^{2} - \frac{1}{6}$

$y = f \left(x\right) = {\left(\frac{3 {x}^{2} - 1}{2}\right)}^{1 / 3}$