Question

\(\;{\bf{9}} - {\bf{14}}\) Find the solution of the differential equation that satisfies the given initial condition.

\(\frac{{dy}}{{dx}} = \frac{x}{y},\;\;\;y(0) = - 3\)

Short answer

\(y = - \sqrt {{x^2} + 9} {\rm{. }}\)

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables

Consider the differential equation\(\frac{{dy}}{{dx}} = \frac{x}{y}\)with initial condition\(y(0) = - 3\).

Separating variable we have

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{x}{y}\\ydy &= xdx\end{aligned}\)

Integrate

Integrating we have:

\(\begin{aligned}\int y dy &= \int x dx + C\\\frac{{{y^{1 + 1}}}}{{1 + 1}} &= \frac{{{x^{1 + 1}}}}{{1 + 1}} + C\;\;\;{\rm{ Using formula }}\int {{x^n}} dx &= \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1\\\frac{1}{2}{y^2} &= \frac{1}{2}{x^2} + C\end{aligned}\)

Apply initial condition

To find value of\(C\),

Using initial condition \(y(0) = - 3\)that means\(x = 0,y = - 3\)

\(\begin{aligned}\frac{1}{2}{( - 3)^2} &= \frac{1}{2}{(0)^2} + C\\\frac{9}{2} &= 0 + C\\\frac{9}{2} &= C\end{aligned}\)

Find solution

Substituting value\(C = \frac{9}{2}\)in\(\frac{1}{2}{y^2} = \frac{1}{2}{x^2} + C\)we have:

\(\begin{aligned}\frac{1}{2}{y^2} &= \frac{1}{2}{x^2} + \frac{9}{2}\\\frac{1}{2}{y^2} &= \frac{1}{2}\left( {{x^2} + 9} \right)\\{y^2} &= {x^2} + 9\\y &= \pm \sqrt {{x^2} + 9} \end{aligned}\)

Since, \(y(0) = - 3 < 0\)we have\(y = - \sqrt {{x^2} + 9} \)

Therefore, the solution of given the differential equations is\(y = - \sqrt {{x^2} + 9} {\rm{. }}\)