Question

Let y=y(x) be the solution of the differential equation ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{dy}{dx}+2x\left( {{x}^{2}}+1 \right)y=1$ such that y(0)=0. If $\sqrt{a}y\left( 1 \right)=\dfrac{\pi }{32}$ then the value of a is:
(a) $\dfrac{1}{2}$
(b) $\dfrac{1}{16}$
(c) $\dfrac{1}{4}$
(d) 1

Answer 1

First, before proceeding for this, we must know the differential equation of the form $\dfrac{dy}{dx}+Py=Q$can be solved easily by using the integrating factor. Then, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as$IF={{e}^{\int{Pdx}}}$. Then, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we have the form of the solution as $y\times IF=\int{Q\times IF}dx+c$. Then, by using the two conditions given in the question, we get the value of a.

Complete step-by-step solution:
In this question, we are supposed to find the value of a for the differential equation as ${{\left( {{x}^{2}}+1 \right)}^{2}}\dfrac{dy}{dx}+2x\left( {{x}^{2}}+1 \right)y=1$ where y(0)=0 and $\sqrt{a}y\left( 1 \right)=\dfrac{\pi }{32}$.
So, before proceeding for this, we must know the differential equation of the form $\dfrac{dy}{dx}+Py=Q$can be solved easily by using the integrating factor.
Now, to make the above equation in the form $\dfrac{dy}{dx}+Py=Q$, dividing both sides by ${{\left( {{x}^{2}}+1 \right)}^{2}}$, we get:
$\dfrac{dy}{dx}+\dfrac{2x}{{{x}^{2}}+1}y=\dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}$
Now, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we need a integrating factor(IF) given by the formula as:
$IF={{e}^{\int{Pdx}}}$
So, the value of P from the above differential equation is $\dfrac{2x}{{{x}^{2}}+1}$ to get the value of IF as:
$IF={{e}^{\int{\dfrac{2x}{{{x}^{2}}+1}dx}}}$
Now, by substituting ${{x}^{2}}+1=t$and by differentiating both sides, we get:
$2xdx=dt$
Then, by substituting all the values in terms of t, we get:
$\begin{aligned} & IF={{e}^{\int{\dfrac{1}{t}dt}}} \\\ & \Rightarrow IF={{e}^{\log t}} \\\ & \Rightarrow IF=t \\\ \end{aligned}$
Then, by again replacing t with its assumed value, we get the If as:
$IF={{x}^{2}}+1$
Now, to get the solution of the above differential equation in the form $\dfrac{dy}{dx}+Py=Q$, we have the form of solution as:
$y\times IF=\int{Q\times IF}dx+c$
Then, by substituting the value of IF and Q, we get:
$\begin{aligned} & y\times \left( {{x}^{2}}+1 \right)=\int{\left( {{x}^{2}}+1 \right)\times \dfrac{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}}}dx+c \\\ & \Rightarrow y\times \left( {{x}^{2}}+1 \right)=\int{\dfrac{1}{\left( {{x}^{2}}+1 \right)}}dx+c \\\ & \Rightarrow y\times \left( {{x}^{2}}+1 \right)={{\tan }^{-1}}x+c \\\ \end{aligned}$
Now, we are given with condition as y(0)=0 which means y is 0 when x is 0 which gives the value of c as:
$\begin{aligned} & 0\times \left( {{0}^{2}}+1 \right)={{\tan }^{-1}}0+c \\\ & \Rightarrow c=0 \\\ \end{aligned}$
So, after substituting the value of c as 0, we get the solution of the differential equation as:
$y\times \left( {{x}^{2}}+1 \right)={{\tan }^{-1}}x$
Now, we are also given with the condition as $\sqrt{a}y\left( 1 \right)=\dfrac{\pi }{32}$which means at x=1, y is $\dfrac{\pi }{32\sqrt{a}}$, we get:
$\dfrac{\pi }{32\sqrt{a}}\times \left( {{1}^{2}}+1 \right)={{\tan }^{-1}}1$
Then, by solving the above expression, we get the value of a as:
$\begin{aligned} & \dfrac{\pi }{32\sqrt{a}}\times \left( 1+1 \right)=\dfrac{\pi }{4} \\\ & \Rightarrow \dfrac{\pi }{32\sqrt{a}}\times 2=\dfrac{\pi }{4} \\\ & \Rightarrow \dfrac{1}{16\sqrt{a}}=\dfrac{1}{4} \\\ & \Rightarrow \dfrac{1}{4\sqrt{a}}=\dfrac{1}{1} \\\ & \Rightarrow \sqrt{a}=\dfrac{1}{4} \\\ & \Rightarrow a={{\left( \dfrac{1}{4} \right)}^{2}} \\\ & \Rightarrow a=\dfrac{1}{16} \\\ \end{aligned}$
So, we get the value of a as $\dfrac{1}{16}$.
Hence, option (b) is correct.

Note: Now, to solve these types of the questions we need to know some of the basic differentiation and integration formulas to get the solution accurately.
So, some of the required formulas are as:
$\begin{aligned} & \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \\\ & \int{\dfrac{1}{x}dx=\log x} \\\ \end{aligned}$
Moreover, there is also an alternative approach to calculate the value of integral with the form as:
$\int{\dfrac{{f}'\left( x \right)}{f\left( x \right)}dx=f\left( x \right)+c}$ which can be used in this question for calculating the integrating factor.