Question

If $y = y(x)$ is the solution of the differential equation $x\dfrac{{dx}}{{dy}} + 2y = {x^2}$ satisfying $y(1) = 1$ , then $y\left( {\dfrac{1}{2}} \right)$ is equal to
A. $\dfrac{{ - 49}}{{16}}$
B. $\dfrac{{49}}{{16}}$
C. $\dfrac{{45}}{8}$
D. $\dfrac{{ - 45}}{8}$

Answer 1

Hint: To demonstrate that the preceding equation is identical to the differential equation $\dfrac{{dx}}{{dy}} - x = y + 1$, divide both sides of the equation by $dy$. Make use of the fact that the integrating factor and the differential equation's solution have the form of: $yIF = \int {Q(x)IFdx}$. To solve the differential equation, we need to obtain the integrating factor of the given equation.

Formula used: The integrating factor is expressed as $I.F. = {e^{\int {P(x)dx} }}$. And the solution of the differential equation $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$is represented as: $yIF = \int {Q(x)IFdx} $.

Complete Step-by-step Solution:
In the question, the differential equation is given by:
$x\dfrac{{dx}}{{dy}} + 2y = {x^2}$
Divide the above equation by$x$to convert the equation in the form of linear differential equation, then we have:
$\dfrac{{x\dfrac{{dx}}{{dy}}}}{x} + \dfrac{{2y}}{x} = \dfrac{{{x^2}}}{x} \\$
$\Rightarrow \dfrac{{dy}}{{dx}} + \left( {\dfrac{2}{x}} \right)y = x\,\,\,\,\,.....(i) \\$
The integrated factor of the linear differential equation is represented as:
$IF = {e^{\int {P(x)dx} }}\,\,\,\,\,\,....(ii)$
By comparing the equation $(i)$ with the integrated factor of linear differential equation, we have:
$P(x) = \dfrac{2}{x}$ and $Q(x) = x$
Now, substitute the value of $P(x)$in the equation $(ii)$, then:
$IF = {e^{\int {\left( {\dfrac{2}{x}} \right)dx} }} \\$
$\Rightarrow IF = {e^{2\log x}} \\$
$\Rightarrow IF = {x^2} \\$
As we know that the solution of the differential equation of the form $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$is expressed as $yIF = \int {Q(x)IFdx} $, then:
$y({x^2}) = \int {(x)} ({x^2})dx \\$
${x^2}y = \int {{x^3}dx} \\$
Now, apply the power rule of integration $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} $for $n \ne - 1$, we have:
${x^2}y = \dfrac{{{x^{3 + 1}}}}{{3 + 1}} + C \\$
$\Rightarrow {x^2}y = \dfrac{{{x^4}}}{4} + C\,\,\,\,\,....(iii) \\$
Given that $y(1) = 1$, so substitute the value of $x = 1$in the obtained equation, then:
${(1)^2}(y) = \dfrac{{{{(1)}^4}}}{4} + C \\$
$\Rightarrow 1 = \dfrac{1}{4} + C \\$
$\Rightarrow C = \dfrac{3}{4} \\$
Pu the obtained value of $C$in the equation $(iii)$,
${x^2}y = \dfrac{{{x^4}}}{4} + \dfrac{3}{4}$
Substitute the value of $x = \dfrac{1}{2}$in the above equation to determine the value of $y\left( {\dfrac{1}{2}} \right)$, we obtain:
$\left( {\dfrac{1}{2}} \right)y = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^4}}}{4} + \dfrac{3}{4} \\$
$\Rightarrow y\left( {\dfrac{1}{2}} \right) = \dfrac{1}{{16}} + \dfrac{3}{{4\left( {\dfrac{1}{4}} \right)}} \\$
$\Rightarrow y\left( {\dfrac{1}{2}} \right) = \dfrac{{49}}{{16}} \\$
Thus, the correct option is: (B) $\dfrac{{49}}{{16}}$

Note: It should be noted that the first order differential equations can be difficult to solve, however the two methods listed below can make things easier. First is about the method of variation, in which the homogeneous equation on the left is solved using methods of variation of parameters to produce a general answer. To accomplish this, the right-hand side is maintained with zero and adjusting the equation as necessary.
And second is about the integrating method, in which we need to put the above first order differential equation in the proper order: $\dfrac{{dy}}{{dx}} + P(x)y = Q(x)$. Then select the integrating factor in the formula $I.F. = {e^{\int {P(x)dx} }}$. After that, add $I.F.$ to the equation to multiply it.