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.

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.

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

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.

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

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 $I.F$.

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

Latest Math Topics

Latest Math Problems

Email subscription

Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects.
Know more

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.