Question

What is a solution to the differential equation $\dfrac{{dy}}{{dx}} = x{y^2}$ with the particular solution $y\left( 2 \right) = - \dfrac{2}{5}$?

Answer 1

Hint: Using the power rule we solve the problem. The power rule for the integration provides us the formula that allows to integrate any function that can be written as a power as negative and fractional exponents to integrate functions involving roots and reciprocal powers of$x$.
Formula used:
Power rule for differentiation is
$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$;
Power rule for integration is
$ = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$;

Complete step-by-step solution:
The given differential equation is,
$\dfrac{{dy}}{{dx}} = x{y^2}$
Given, $y\left( 2 \right) = - \dfrac{2}{5}$
Separate the $x$and $y$terms we have,
$\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} = x$
Take integral over the equation, we get,
$\int {\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}}dx = \int {xdx} } $
Cancel the term $dx$on the left hand side, we have
$\int {\dfrac{1}{{{y^2}}}dy = \int {xdx} } $ ……………………$\left( 1 \right)$
By using the power rule, integrate the equation$\left( 1 \right)$,
We already known the power rule for integration,
The denominator ${y^2}$comes to the numerator it will becomes ${y^{ - 2}}$
$\int {{y^{ - 2}}dy = \int {xdx} } $
Applying the power rule
$ = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
First we apply the power rule of integration on the left hand side, we have,
$\int {{y^{ - 2}}dy} $
Applying the rule,
$ = \dfrac{{{y^{ - 2 + 1}}}}{{ - 2 + 1}} + C$
$ = \dfrac{{{y^{ - 1}}}}{{ - 1}}$
Now the numerator comes to denominator, we have,
$ = - \dfrac{1}{y}$
Now we apply the power rule of integration on the right hand side. At the right hand side we have,
$\int {xdx} $
Applying the power rule, we get,
$ = \dfrac{{{x^{1 + 1}}}}{{1 + 1}} + C$
$ = \dfrac{{{x^2}}}{2} + C$
Therefore after applying the power rule, we get,
$- \dfrac{1}{y} = \dfrac{{{x^2}}}{2} + C$ ………………………………$\left( 2 \right)$
Given,
$y\left( 2 \right) = - \dfrac{2}{5}$
$\dfrac{5}{2} = 2 + C$
$C = \dfrac{5}{2} - 2$
Take least common multiple to solve this, we get,
$C = \dfrac{{5 - 4}}{2}$
$C = \dfrac{1}{2}$
Substitute the value in the equation $\left( 2 \right)$we get,
$\dfrac{{ - 1}}{y} = \dfrac{{{x^2}}}{2} + c$
$\dfrac{{ - 1}}{y} = \dfrac{{{x^2}}}{2} + \dfrac{1}{2}$
Take $\dfrac{1}{2}$ as a common value, we have,
$\dfrac{{ - 1}}{y} = \dfrac{1}{2}\left( {{x^2} + 1} \right)$
Apply cross multiplication,
$\dfrac{{ - 2}}{y} = \left( {{x^2} + 1} \right)$
Take the reciprocal to this equation, we have,
$\dfrac{{ - y}}{2} = \dfrac{1}{{{x^2} + 1}}$
Solve this equation to find the value of y is,
$y = - \dfrac{2}{{{x^2} + 1}}$
Therefore the solution for the differential equation is,
$y = - \dfrac{2}{{{x^2} + 1}}$;

Note: In calculus, the power rule is used to differentiate functions of the form $f\left( x \right) = {x^r}$whenever $r$ is a real number. Since differentiation is a linear operation on the space of differentiable functions, polynomials can also be differentiated using this rule.