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Question: What is a solution to the differential equation \(\dfrac{{dy}}{{dx}} = x{y^2}\) with the particular ...

What is a solution to the differential equation dydx=xy2\dfrac{{dy}}{{dx}} = x{y^2} with the particular solution y(2)=−25y\left( 2 \right) = - \dfrac{2}{5}?

Explanation

Solution

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 ofxx.
Formula used:
Power rule for differentiation is
ddxxn=nxn−1\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}};
Power rule for integration is
=xn+1n+1+C= \dfrac{{{x^{n + 1}}}}{{n + 1}} + C;

Complete step-by-step solution:
The given differential equation is,
dydx=xy2\dfrac{{dy}}{{dx}} = x{y^2}
Given, y(2)=−25y\left( 2 \right) = - \dfrac{2}{5}
Separate the xxand yyterms we have,
1y2dydx=x\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}} = x
Take integral over the equation, we get,
∫1y2dydxdx=∫xdx\int {\dfrac{1}{{{y^2}}}\dfrac{{dy}}{{dx}}dx = \int {xdx} }
Cancel the term dxdxon the left hand side, we have
∫1y2dy=∫xdx\int {\dfrac{1}{{{y^2}}}dy = \int {xdx} } ……………………(1)\left( 1 \right)
By using the power rule, integrate the equation(1)\left( 1 \right),
We already known the power rule for integration,
The denominator y2{y^2}comes to the numerator it will becomes y−2{y^{ - 2}}
∫y−2dy=∫xdx\int {{y^{ - 2}}dy = \int {xdx} }
Applying the power rule
=xn+1n+1+C= \dfrac{{{x^{n + 1}}}}{{n + 1}} + C
First we apply the power rule of integration on the left hand side, we have,
∫y−2dy\int {{y^{ - 2}}dy}
Applying the rule,
=y−2+1−2+1+C= \dfrac{{{y^{ - 2 + 1}}}}{{ - 2 + 1}} + C
=y−1−1= \dfrac{{{y^{ - 1}}}}{{ - 1}}
Now the numerator comes to denominator, we have,
=−1y= - \dfrac{1}{y}
Now we apply the power rule of integration on the right hand side. At the right hand side we have,
∫xdx\int {xdx}
Applying the power rule, we get,
=x1+11+1+C= \dfrac{{{x^{1 + 1}}}}{{1 + 1}} + C
=x22+C= \dfrac{{{x^2}}}{2} + C
Therefore after applying the power rule, we get,
−1y=x22+C- \dfrac{1}{y} = \dfrac{{{x^2}}}{2} + C ………………………………(2)\left( 2 \right)
Given,
y(2)=−25y\left( 2 \right) = - \dfrac{2}{5}
52=2+C\dfrac{5}{2} = 2 + C
C=52−2C = \dfrac{5}{2} - 2
Take least common multiple to solve this, we get,
C=5−42C = \dfrac{{5 - 4}}{2}
C=12C = \dfrac{1}{2}
Substitute the value in the equation (2)\left( 2 \right)we get,
−1y=x22+c\dfrac{{ - 1}}{y} = \dfrac{{{x^2}}}{2} + c
−1y=x22+12\dfrac{{ - 1}}{y} = \dfrac{{{x^2}}}{2} + \dfrac{1}{2}
Take 12\dfrac{1}{2} as a common value, we have,
−1y=12(x2+1)\dfrac{{ - 1}}{y} = \dfrac{1}{2}\left( {{x^2} + 1} \right)
Apply cross multiplication,
−2y=(x2+1)\dfrac{{ - 2}}{y} = \left( {{x^2} + 1} \right)
Take the reciprocal to this equation, we have,
−y2=1x2+1\dfrac{{ - y}}{2} = \dfrac{1}{{{x^2} + 1}}
Solve this equation to find the value of y is,
y=−2x2+1y = - \dfrac{2}{{{x^2} + 1}}
Therefore the solution for the differential equation is,
y=−2x2+1y = - \dfrac{2}{{{x^2} + 1}};

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