Question

Verify that $y=C{x}^3$ is a solution of the differential equation $x{y}^{\prime }-3y=0$ for any value of C. Then
find the particular solution determined by the initial condition $y=2$ when $x=-3$.

• A. $y=\frac{2}{27}{x}^2$
• B. $y=-\frac{2}{27}{x}^3$
• C. $y=-\frac{2}{25}{x}^3$
• D. None of these

Answer 1

Answer: (B)

$y=-\frac{2}{27}{x}^3$

Given that $y=C{x}^3$ is the solution of the differential equation$x{y}^{{}^{\prime }}=3y$

So to get a particular solution we need to find the value of $C$ for a given initial value condition,

i.e.,$y=2$ when $x=-3$,

Substituting this in the general solution gives,

$2=C(-27)$

$⟹C=-\frac{2}{27}$

$\therefore$ the particular solution is $y=-\frac{2}{27}{x}^3$