Question

Solve the differential equation $\left( 1+{{x}^{2}} \right)\dfrac{dy}{dx}+y={{e}^{{{\tan }^{-1}}x}}$.

Answer 1

Hint: First convert the equation to the form $\dfrac{dy}{dx}+p\left( x \right)y=g\left( x \right)$ and find the integration factor of the differential equation using $IF={{e}^{\int{p\left( x \right)dx}}}$ . Then the solution of the differential equation is $y\left( IF \right)=\int{\left( IF \right)g\left( x \right)dx}$ . just solve the integral on the right-hand side of the equation and report the answer. To solve integral, the method of substitution is to be used, by taking $t={{\tan }^{-1}}x$.

Complete step by step answer:
Let us start the solution to the above question by converting the equation to the form $\dfrac{dy}{dx}+p\left( x \right)y=g\left( x \right)$ .
$\left( 1+{{x}^{2}} \right)\dfrac{dy}{dx}+y={{e}^{{{\tan }^{-1}}x}}$
$\Rightarrow \dfrac{dy}{dx}+\dfrac{y}{1+{{x}^{2}}}=\dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$
Now, we know that the integration factor of the differential equation $\dfrac{dy}{dx}+p\left( x \right)y=g\left( x \right)$ is $IF={{e}^{\int{p\left( x \right)dx}}}$ . For our equation $p\left( x \right)=\dfrac{1}{1+{{x}^{2}}}$ . So, the IF of our differential equation is:
$IF={{e}^{\int{\dfrac{dx}{1+{{x}^{2}}}}}}={{e}^{{{\tan }^{-1}}x}}$
So, the solution to the differential equation $\dfrac{dy}{dx}+p\left( x \right)y=g\left( x \right)$ is: $y\left( IF \right)=\int{\left( IF \right)g\left( x \right)dx}$ .
So, the solution to our differential equation is:
$y{{e}^{{{\tan }^{-1}}x}}=\int{{{e}^{{{\tan }^{-1}}x}}\times \dfrac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}$
$\Rightarrow y{{e}^{{{\tan }^{-1}}x}}=\int{\dfrac{{{e}^{2{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}$
Now, to solve the integral, we will let ${{\tan }^{-1}}x$ to be t.
$t={{\tan }^{-1}}x$
Now, we will differentiate both sides of the equation. On doing so, we get
$\dfrac{dt}{dx}=\dfrac{d({{\tan }^{-1}}x)}{dx}$
$\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{1+{{x}^{2}}}$
$\Rightarrow dt=\dfrac{dx}{1+{{x}^{2}}}$
Now, we will substitute t and dt in our integral. On doing so, we get
\[y{{e}^{{{\tan }^{-1}}x}}=\int{{{e}^{2t}}dt}\]
Now, we will use the identity that $\int{{{e}^{kx}}dx}=\dfrac{{{e}^{kx}}}{k}+c$ , where k is constant.
\[y{{e}^{{{\tan }^{-1}}x}}=\dfrac{{{e}^{2t}}}{2}+c\]
Now, we will again substitute the value of t according to our assumption. On doing so, we get
\[y{{e}^{{{\tan }^{-1}}x}}=\dfrac{{{e}^{2{{\tan }^{-1}}x}}}{2}+c\]

Hence, the answer to the above question is \[y{{e}^{{{\tan }^{-1}}x}}=\dfrac{{{e}^{2{{\tan }^{-1}}x}}}{2}+c\].

Note: Remember that integrating factor and the above used method is only valid for the differential equations which can be written in the form of $\dfrac{dy}{dx}+p\left( x \right)y=g\left( x \right)$ . Also, don’t forget the constant of integration or to substitute the assumed variables while reporting the final answer. Don’t make the mistake and try to substitute $t=\dfrac{1}{1+{{x}^{2}}}$ to solve the integral, as it won’t give any useful results.