Question

If y = y (x) is the solution of the differential equation, $x\dfrac{dy}{dx}+2y={{x}^{2}}$ satisfying y (1) = 1, then $y\left( \dfrac{1}{2} \right)$ is equal to?
A.$\dfrac{7}{64}$
B.$\dfrac{13}{16}$
C.$\dfrac{49}{16}$
D.$\dfrac{1}{4}$

Answer 1

Hint:Convert the given equation in the $\dfrac{dy}{dx}+Py=Q$ and then use the formula $y\times I.F.=\int{\left( Q\times I.F. \right)dx}+C$ where, $I.F.={{e}^{\int{Pdx}}}$ to get the general solution. Then use the condition y (1) = 1 to find the value of C, and finally put \[x=\dfrac{1}{2}\] in the general solution to get the value of $y\left( \dfrac{1}{2} \right)$.

Complete step by step answer:
To find the value of $y\left( \dfrac{1}{2} \right)$ we will write down the given equation first, therefore,
$x\dfrac{dy}{dx}+2y={{x}^{2}}$
If we divide the above equation by ‘x’ we will get,
$\therefore \dfrac{x\dfrac{dy}{dx}+2y}{x}=\dfrac{{{x}^{2}}}{x}$
Separating the denominator in the above equation we will get,
$\therefore \dfrac{dy}{dx}+\dfrac{2y}{x}=x$ …………………………………………………………………………. (1)
If we compare the above equation with $\dfrac{dy}{dx}+Py=Q$ we will get,
$P=\dfrac{2}{x}$ And Q = x ……………………………………………………. (2)
As given equation is of the form given by $\dfrac{dy}{dx}+Py=Q$ therefore we should know the formula to find the general solution for it which is given below,
Formula:
If the differential equation is of the form $\dfrac{dy}{dx}+Py=Q$ then its general solution is given by,
$y\times I.F.=\int{\left( Q\times I.F. \right)dx}+C$ where, $I.F.={{e}^{\int{Pdx}}}$ …………………………………………………… (3)
Before using the formula of general solution we will first calculate the value of Integrating Factor i.e. I.F., Therefore we will get,
$I.F.={{e}^{\int{Pdx}}}$
If we put the value of P from equation (2) in the above equation we will get,
\[\therefore I.F.={{e}^{\int{\dfrac{2}{x}dx}}}\]
By taking constant outside from integration we will get,
\[\therefore I.F.={{e}^{2\int{\dfrac{1}{x}dx}}}\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[\int{\dfrac{1}{x}dx}=\log x\]
By using the formula given above we can write I.F. as,
\[\therefore I.F.={{e}^{2\log x}}\]
As we know that \[m\log n=\log {{m}^{n}}\] therefore above equation will become,
\[\therefore I.F.={{e}^{\log {{x}^{2}}}}\]
As we know that \[{{e}^{{{\log }_{e}}m}}=m\] by using this concept in the above equation we will get,
\[\therefore I.F.={{x}^{2}}\] ……………………………………………………………… (4)
Now we will write the formula of general solution from equation (3) we will get,
$y\times I.F.=\int{\left( Q\times I.F. \right)dx}+C$
If we put the values of equation (2) and equation (4) in the above equation we will get,
\[y\times {{x}^{2}}=\int{\left( x\times {{x}^{2}} \right)dx}+C\]
If we simplify the above equation we will get,
\[\therefore {{x}^{2}}y=\int{\left( {{x}^{3}} \right)dx}+C\]
To proceed further in the solution we should know the formula given below,
Formula:
\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+2}\]
By using the above formula in our general solution we will get,
\[\therefore {{x}^{2}}y=\dfrac{{{x}^{3+1}}}{3+1}+C\]
Further simplification in the above solution will give,
\[\therefore {{x}^{2}}y=\dfrac{{{x}^{4}}}{4}+C\] ……………………………………………………………………… (5)
As we have given in the question that y (1) = 1 i.e. if we put x = 1 in the general solution then the value of ‘y’ will become 1 i.e. y = 1 and if we put these values in the above equation we will get,
\[\therefore {{1}^{2}}\times 1=\dfrac{{{1}^{4}}}{4}+C\]
Further simplification in the above equation will give,
\[\therefore 1=\dfrac{1}{4}+C\]
\[\therefore -C=\dfrac{1}{4}-1\]
\[\therefore -C=\dfrac{1-4}{4}\]
\[\therefore -C=-\dfrac{3}{4}\]
\[\therefore C=\dfrac{3}{4}\]
Now if we put the value of C in equation (5) we will get our general solution as,
\[\therefore {{x}^{2}}y=\dfrac{{{x}^{4}}}{4}+\dfrac{3}{4}\]
As we have to find the value of $y\left( \dfrac{1}{2} \right)$ therefore we will put \[x=\dfrac{1}{2}\] in the above equation, therefore we will get,
\[\therefore {{\left( \dfrac{1}{2} \right)}^{2}}y=\dfrac{{{\left( \dfrac{1}{2} \right)}^{4}}}{4}+\dfrac{3}{4}\]
\[\therefore \dfrac{1}{4}y=\dfrac{\dfrac{1}{16}}{4}+\dfrac{3}{4}\]
\[\therefore \dfrac{1}{4}y=\dfrac{1}{16}\times \dfrac{1}{4}+\dfrac{3}{4}\]
If we multiply the above equation by 4 we will get,
\[\therefore 4\times \dfrac{1}{4}y=4\times \left( \dfrac{1}{16}\times \dfrac{1}{4}+\dfrac{3}{4} \right)\]
\[\therefore y=\dfrac{1}{16}+3\]
\[\therefore y=\dfrac{1+16\times 3}{16}\]
\[\therefore y=\dfrac{1+48}{16}\]
\[\therefore y=\dfrac{49}{16}\]
Therefore the value of \[y\left( \dfrac{1}{2} \right)\] is equal to \[\dfrac{49}{16}\].
Therefore the correct answer is option (c).

Note: While calculating the I.F. always remember that the base of ‘log’ is always ‘e’ and not ‘10’ if the solution of integration comes in the logarithmic form. If you take it as 10 then you will complicate the solution and the solution will also become wrong.
Like, \[I.F.={{e}^{{{\log }_{e}}{{x}^{2}}}}\].