Question

Solve the differential equation $\dfrac{dy}{dx}-\dfrac{2xy}{1+{{x}^{2}}}=1+{{x}^{2}}$.

Answer 1

To solve this differential equation, we should know the concept of linear differential equation of first order. We can infer from the given differential equation that its order is 1. The solution of a differential equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ is given by $y{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right)}{{e}^{\int{P\left( x \right)dx}}}dx+c$. The term ${{e}^{\int{P\left( x \right)dx}}}$ is known as the integration factor I.F of the differential equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. Using this formula, by substituting $P\left( x \right)=-\dfrac{2x}{1+{{x}^{2}}}$ and $Q\left( x \right)={{x}^{2}}+1$, we get the required solution of the differential equation.

Complete step-by-step answer:
Let us consider the equation $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. We don’t have any direct method to solve this equation. To solve this equation, we should multiply the whole equation by another function of x. Let the function be $R\left( x \right)$. Multiplying by $R\left( x \right)$, we get
$R\left( x \right)\dfrac{dy}{dx}+R\left( x \right).P\left( x \right)y=Q\left( x \right).R\left( x \right)$
The L.H.S can be modified into a derivative of one function by assuming a relation between $R\left( x \right)$ and $P\left( x \right)$. The relation is
$\dfrac{d\left( R\left( x \right) \right)}{dx}=R\left( x \right).P\left( x \right)\to \left( 1 \right)$
Using equation-1 we get the differential equation as
$R\left( x \right)\dfrac{dy}{dx}+\dfrac{d\left( R\left( x \right) \right)}{dx}y=Q\left( x \right).R\left( x \right)$
The L.H.S is in the form of the product rule in differentiation which is
$\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$
Using this rule, we can change the equation as

& \dfrac{d\left( R\left( x \right).y \right)}{dx}=Q\left( x \right).R\left( x \right) \\\ & d\left( R\left( x \right).y \right)=Q\left( x \right).R\left( x \right)dx \\\ \end{aligned}$$ Integrating on both sides, we get $$\begin{aligned} & \int{d\left( R\left( x \right).y \right)}=\int{Q\left( x \right).R\left( x \right)dx}+c \\\ & R\left( x \right).y=\int{Q\left( x \right).R\left( x \right)dx}+c\to \left( 2 \right) \\\ \end{aligned}$$ Now let us come back to the equation-1 $\begin{aligned} & \dfrac{d\left( R\left( x \right) \right)}{dx}=R\left( x \right).P\left( x \right) \\\ & \dfrac{d\left( R\left( x \right) \right)}{R\left( x \right)}=P\left( x \right)dx \\\ \end{aligned}$ We know that $\int{\dfrac{1}{f\left( x \right)}d\left( f\left( x \right) \right)}=\ln \left( f\left( x \right) \right)$ Using this and integrating the above equation, we get $\begin{aligned} & \int{\dfrac{d\left( R\left( x \right) \right)}{R\left( x \right)}}=\int{P\left( x \right)dx} \\\ & \ln \left( R\left( x \right) \right)=\int{P\left( x \right)dx} \\\ \end{aligned}$ We know that $\ln x=a\Rightarrow x={{e}^{a}}$, using this, we get $R\left( x \right)={{e}^{\int{P\left( x \right)dx}}}$ Using this expression in equation-2, we get $$y.{{e}^{\int{P\left( x \right)dx}}}=\int{Q\left( x \right).{{e}^{\int{P\left( x \right)dx}}}dx}+c$$ In our question $P\left( x \right)=-\dfrac{2x}{1+{{x}^{2}}}$ and $Q\left( x \right)={{x}^{2}}+1$. Let us find the value of ${{e}^{\int{P\left( x \right)dx}}}$ ${{e}^{\int{P\left( x \right)dx}}}={{e}^{\int{\dfrac{-2x}{1+{{x}^{2}}}dx}}}$ We know that $2xdx=d\left( {{x}^{2}} \right)$. Using this in above equation, we get ${{e}^{\int{\dfrac{-2x}{1+{{x}^{2}}}dx}}}={{e}^{-\int{\dfrac{1}{1+{{x}^{2}}}d\left( {{x}^{2}} \right)}}}$ Using the relation $\int{\dfrac{1}{f\left( x \right)}d\left( f\left( x \right) \right)}=\ln \left( f\left( x \right) \right)$, we can write that ${{e}^{-\int{\dfrac{1}{1+{{x}^{2}}}d\left( {{x}^{2}} \right)}}}={{e}^{-\ln \left( 1+{{x}^{2}} \right)}}$ We know that ${{e}^{-\ln x}}=\dfrac{1}{x}$, we can write that ${{e}^{-\int{\dfrac{1}{1+{{x}^{2}}}d\left( {{x}^{2}} \right)}}}={{e}^{-\ln \left( 1+{{x}^{2}} \right)}}=\dfrac{1}{1+{{x}^{2}}}$ So, ${{e}^{\int{P\left( x \right)dx}}}=\dfrac{1}{1+{{x}^{2}}}$ Using this, we get the solution as $$\begin{aligned} & y.\dfrac{1}{1+{{x}^{2}}}=\int{\left( {{x}^{2}}+1 \right).\dfrac{1}{1+{{x}^{2}}}dx}+c \\\ & \dfrac{y}{1+{{x}^{2}}}=\int{1dx}+c \\\ & \dfrac{y}{1+{{x}^{2}}}=x+c \\\ & y=x\left( {{x}^{2}}+1 \right)+c\left( {{x}^{2}}+1 \right) \\\ \end{aligned}$$ $\therefore $The solution of the given differential equation is $$y=x\left( {{x}^{2}}+1 \right)+c\left( {{x}^{2}}+1 \right)$$ **Note:** We can check whether our solution is correct or not in these kinds of questions related to differential equations. Let us differentiate y with respect to x. We get $$\dfrac{dy}{dx}=3{{x}^{2}}+1+2cx$$ Substituting the values of y and $\dfrac{dy}{dx}$ in the L.H.S of differential equation, we get $$\begin{aligned} & 3{{x}^{2}}+1+2cx-\dfrac{2x}{1+{{x}^{2}}}\left( x\left( {{x}^{2}}+1 \right)+c\left( {{x}^{2}}+1 \right) \right) \\\ & =3{{x}^{2}}+1+2cx-2x\left( x+c \right)=3{{x}^{2}}+1+2cx-2{{x}^{2}}-2cx={{x}^{2}}+1=R.H.S \\\ \end{aligned}$$ Hence the solution $$y=x\left( {{x}^{2}}+1 \right)+c\left( {{x}^{2}}+1 \right)$$ satisfies the differential equation and the result is verified.