Math Doubts

Solution for Leibniz’s Linear differential equation

A German mathematician Gottfried Wilhelm Leibnitz (or Leibniz) defined a linear differential equation of first order in standard form in terms of two functions $P$ and $Q$. The functions $P$ and $Q$ are two functions in terms of $x$.

$\dfrac{dy}{dx}+Py$ $\,=\,$ $Q$

Now, let us learn how to find the solution for the Leibnitz’s linear differential equation.

Multiply the equation by an integrating factor

Gottfried Wilhelm Leibniz introduced the solution for his standard form linear differential equation by multiplying it with an integrating factor $e$ raised to the power the integral of $P$ with respect to $x$. So, multiply the expressions on both sides of the equation with this integrating factor.

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \bigg(\dfrac{dy}{dx}+Py\bigg)$ $\,=\,$ $e^{\displaystyle \int{P}\,dx} \times Q$

The integrating factor can be distributed over the sum of two terms by using the distributive property of multiplication across the addition.

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $e^{\displaystyle \int{P}\,dx} \times Py$ $\,=\,$ $e^{\displaystyle \int{P}\,dx} \times Q$

Prepare the Equation for Integration by simplification

There is nothing to simplify in the expression on right side of the equation. So, let’s only concentrate on simplifying the binomial on the left-hand side of the equation.

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $e^{\displaystyle \int{P}\,dx} \times P \times y$ $\,=\,$ $Q \times e^{\displaystyle \int{P}\,dx}$

In the first term, the integrating factor $e$ raised to the power the integral of $P$ with respect to $x$ is multiplied by the derivative of $y$ with respect to $x$. In the second term, the variable $y$ is multiplied by the product of integral factor and the function $P$ in terms of $x$.

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $\bigg(e^{\displaystyle \int{P}\,dx} \times P\bigg) \times y$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $y \times \bigg(e^{\displaystyle \int{P}\,dx} \times P\bigg)$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $y \times \bigg(e^{\displaystyle \int{P}\,dx} P\bigg)$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

The derivative of the integrating factor $e$ raised to the power of integral of $P$ with respect to $x$ is equal to the product of the integrating factor and the function $P$ in terms of $x$.

$\dfrac{d}{dx}{\bigg(e^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $e^{\displaystyle \int{P}\,dx} P$

Therefore, the product of integral factor and the function $P$ in terms of $x$ can be written as the derivative of the integrating factor in the second term.

$\implies$ $e^{\displaystyle \int{P}\,dx} \times \dfrac{dy}{dx}$ $+$ $y \times \dfrac{d}{dx}{\bigg(e^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

The sum of the terms represents the derivative of the product of the variable $y$ and the integrating factor $e$ raised to the power of the integral of the function $P$ with respect to $x$. Hence, the binomial can be simplified by using the product rule of the derivatives.

$\implies$ $\dfrac{d}{dx}{\bigg(e^{\displaystyle \int{P}\,dx} \times y\bigg)}$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

$\implies$ $\dfrac{d}{dx}{\bigg(y \times e^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

$\implies$ $\dfrac{d}{dx}{\bigg(ye^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx}$

The simplification procedure of the Leibnitz’s linear differential equation has reached to the final stage. So, shift the differential $dx$ to the right-hand side of the equation.

$\implies$ $d{\bigg(ye^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $Qe^{\displaystyle \int{P}\,dx} \times dx$

Solve the Linear differential equation by integration

Now, take the indefinite integral on both sides of the equation to solve the linear differential equation of first order.

$\implies$ $\displaystyle \int{}d{\bigg(ye^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $\displaystyle \int{Qe^{\displaystyle \int{P}\,dx}}\,dx$

$\implies$ $\displaystyle \int{1 \times}d{\bigg(ye^{\displaystyle \int{P}\,dx}\bigg)}$ $\,=\,$ $\displaystyle \int{Qe^{\displaystyle \int{P}\,dx}}\,dx$

On the left hand side of the equation, the integral of $1$ with respect to the product of $y$ and the integrating factor is equal to the product of the variable $y$ and the integrating factor $e$ raised to the power of the integral of function $P$ with respect to $x$ as per the integral rule of one.

$\implies$ $ye^{\displaystyle \int{P}\,dx}$ $+$ $c_1$ $\,=\,$ $\displaystyle \int{Qe^{\displaystyle \int{P}\,dx}}\,dx$ $+$ $c_2$

Here, $c_1$ is the constant of integration for the left hand side of the equation and $c_2$ is the integral constant after the integration for the right hand side expression.

$\implies$ $ye^{\displaystyle \int{P}\,dx}$ $\,=\,$ $\displaystyle \int{Qe^{\displaystyle \int{P}\,dx}}\,dx$ $+$ $c_2$ $-$ $c_1$

The difference of the constants is also a constant. Hence, it is simply denoted by a constant $c$.

$\,\,\,\therefore\,\,\,\,\,\,$ $ye^{\displaystyle \int{P}\,dx}$ $\,=\,$ $\displaystyle \int{Qe^{\displaystyle \int{P}\,dx}}\,dx$ $+$ $c$

It is the solution for the Leibniz’s (or Leibnitz’s) standard form linear differential equation.

For our convenience, the integrating factor $e$ raised to the power the integral of $P$ with respect to $x$ is simply denoted by $I.F$.

$\,\,\,\therefore\,\,\,\,\,\,$ $y(I.F)$ $\,=\,$ $\displaystyle \int{Q(I.F)}\,dx$ $+$ $c$

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