Question

How do you find the general solution of \[\dfrac{{dy}}{{dx}} = \dfrac{x}{1 + x^{2}}\] ?

Answer 1

Hint: In this question, we need to find the general solution of the differential equation \[\dfrac{{dy}}{{dx}} = \dfrac{x}{1 + x^{2}}\] . We can find the general solution of the differential equation by using separation of variable methods. First by using the separation of variables method, we can separate the derivative term and variable term. Then we need to use the reverse chain rule by substituting the function as \[u\] . On further integrating, we can find the integral of the given expression.

Formula used :
1. \[\dfrac{d}{{dx}}\left( x^{n} \right) = nx^{n – 1}\]
2. \[\dfrac{d}{{dx}}\left( k \right) = 0\]
3. \[m\ ln(n) = ln\left( n \right)^{m}\]

Complete step-by-step answer:
Given,
\[\dfrac{dy}{dx} = \dfrac{x}{1 + x^{2}}\]
The degree and order of the given differential equation is one. Thus we can use the separation of variables method to find the general solution of the given equation.
In the separation of variables method, we need to separate the derivative term and variable term to another side of equality.
We have,
\[\dfrac{dy}{dx} = \dfrac{x}{1 + x^{2}}\]
For the separation of variables, we need to multiply both sides by \[{dx}\] ,
On multiplying both sides by \[{dx}\],
We get,
\[dy = \dfrac{x}{1 + x^{2}}{dx}\]
On integrating both sides,
We get,
\[\int dy = \int\left( \dfrac{x}{1 + x^{2}} \right){dx}\] ••• (1)
Let us consider \[u = \left( 1 + x^{2} \right)\]
On differentiating \[u\] ,
We get,
\[\dfrac{du}{dx} = 2x + 0\]
\[\Rightarrow \dfrac{du}{2} = x\ dx\]
Thus equation (1) becomes,
\[\int dy = \int\left( \dfrac{1}{u} \right)\left( \dfrac{du}{2} \right)\]
By rewriting the terms,
We get,
\[\int dy = \dfrac{1}{2}\int\left( \dfrac{1}{u} \right){du}\]
We know that \[\int\dfrac{1}{x}dx = ln\left( x \right) + \ c\]
Thus by using reciprocal rule,
We get,
\[\Rightarrow \ y = \dfrac{1}{2}\ln\left( u \right) + c\]
Where \[c\] is the constant of integration.
By substituting the value of \[u\],
We get,
\[y = \dfrac{1}{2}\ln\left( x^{2} + 1 \right) + c\]
We know that \[m\ ln(n) = ln\left( n \right)^{m}\]
From the logarithmic rule,
We get,
\[y = ln\left( x^{2} + 1 \right)^{\dfrac{1}{2}} + c\]
\[\Rightarrow \ y = \ln\left( \sqrt{x^{2} + 1} \right) + c\]
Thus the general solution of \[\dfrac{dy}{dx} = \dfrac{x}{1 + x^{2}}\] is \[y = ln\left( \sqrt{x^{2} + 1} \right) + c\]
Final answer :
The general solution of \[\dfrac{dy}{dx} = \dfrac{x}{1 + x^{2}}\] is \[y = ln\left( \sqrt{x^{2} + 1} \right) + c\]

Note: The concept used in this question is integration method, that is integration by separation of variable methods , u-substitution and reciprocal rule . Since this is an indefinite integral we have to add an arbitrary constant ‘\[c\]’. \[c\] is called the constant of integration. The variable \[x\] in \[dx\] is known as the variable of integration or integrator. Mathematically, integrals are also used to find many useful quantities such as areas, volumes, displacement, etc. The reverse chain rule method is related to the chain rule of differentiation, which when applied to antiderivatives is known as the reverse chain rule that is integration by \[u\] substitution.