Question

The solution of the differential equation \[x\dfrac{dy}{dx}+2y={{x}^{2}}\left( x\ne 0 \right)\] with \[y\left( 1 \right)=1,\] is:
\[\left( a \right)y=\dfrac{{{x}^{3}}}{5}+\dfrac{1}{5{{x}^{2}}}\]
\[\left( b \right)y=\dfrac{4}{5}{{x}^{3}}+\dfrac{1}{5{{x}^{2}}}\]
\[\left( c \right)y=\dfrac{3}{4}{{x}^{2}}+\dfrac{1}{4{{x}^{2}}}\]
\[\left( d \right)y=\dfrac{{{x}^{2}}}{4}+\dfrac{3}{4{{x}^{2}}}\]

Answer 1

Hint: We are given the differential equation as \[x\dfrac{dy}{dx}+2y={{x}^{2}}.\] We will convert it into a linear form \[\dfrac{dy}{dx}+Py=Q\left( x \right)\] and then find the integrating factor using \[I.F={{e}^{\int{Pdx}}}.\] Once we have our integrating factor, we will find the solution using \[y\times IF=\int{\left( Q\times IF \right)dx+C}.\] We will put the value of Q and IF and solve for y which is our required solution.

Complete step-by-step answer:
We are given the differential equation as \[x\dfrac{dy}{dx}+2y={{x}^{2}}.\] First, we will convert our differential equation into the linear form. We have
\[x\dfrac{dy}{dx}+2y={{x}^{2}}\]
Dividing both the sides by x, we get,
\[\Rightarrow \dfrac{dy}{dx}+\dfrac{2y}{x}=x\]
So, we get our linear form. So comparing with \[\dfrac{dy}{dx}+Py=Q,\] we get, \[P=\dfrac{2}{x}\] and Q = x.
Now, to solve further we will find the integrating factor. The integrating factor is given as
\[I.F={{e}^{\int{Pdx}}}\]
As, \[P=\dfrac{2}{x},\] so we get,
\[\Rightarrow I.F={{e}^{\int{\dfrac{2}{x}dx}}}\]
As, \[\int{\dfrac{1}{x}dx}=\log x,\] so we get,
\[\Rightarrow IF={{e}^{2\log x}}\]
As, \[n\log x=\log {{x}^{n}},\] so we get,
\[\Rightarrow IF={{e}^{\log {{x}^{2}}}}\]
Now, we know that, \[{{e}^{\log x}}=x,{{e}^{\log {{x}^{2}}}}={{x}^{2}},\] so we get,
\[\Rightarrow IF={{x}^{2}}\]
Now, we know the solution is given as,
\[y\times IF=\int{\left( Q\times IF \right)dx+C}\]
As, \[IF={{x}^{2}},Q=x,\] so we get,
\[\Rightarrow y{{x}^{2}}=\int{\left( x\times {{x}^{2}} \right)dx}\]
\[\Rightarrow y{{x}^{2}}=\int{{{x}^{3}}dx}\]
As \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n},\] so,
\[\Rightarrow y{{x}^{2}}=\dfrac{{{x}^{4}}}{4}+C\]
Dividing both the sides by \[{{x}^{2}},\] we get,
\[\Rightarrow y=\dfrac{{{x}^{2}}}{4}+\dfrac{C}{{{x}^{2}}}\]
Now, we have y(1) as 1, i.e y (1) = 1. So, putting x = 1 and y = 1, we get,
\[\Rightarrow 1=\dfrac{{{1}^{2}}}{4}+\dfrac{C}{{{1}^{2}}}\]
\[\Rightarrow 1=\dfrac{1}{4}+\dfrac{C}{1}\]
\[\Rightarrow 1=\dfrac{1}{4}+C\]
Solving for C, we get,
\[\Rightarrow C=1-\dfrac{1}{4}=\dfrac{3}{4}\]
Now, we get, \[C=\dfrac{3}{4}.\]
So, putting it in the value of y, we get,
\[\Rightarrow y=\dfrac{{{x}^{2}}}{4}+\dfrac{3}{4{{x}^{2}}}\]

So, the correct answer is “Option d”.

Note: Remember that while finding the integrating factor, we apply the differentiation. First, we integrate P with respect to x and then solve it over the exponential. As we do that we have P as \[\dfrac{2}{x}.\] So,
\[\int{\dfrac{2}{x}dx}=2\int{\dfrac{1}{x}dx}\]
\[=2\log x\]
Then we apply \[{{e}^{\int{\dfrac{2}{x}dx}}}={{e}^{2\log x}}.\] Also, \[n\log x=\log {{x}^{n}}.\] So, we get, \[{{e}^{\log {{x}^{2}}}}\] and then we simplify and get IF as \[{{x}^{2}}.\] So, we do it by part by part so that no errors occur.