Maths > Differential Equations > 2.0 Methods to find the solution of first order and first degree differential equation
Differential Equations
1.0 Introduction
2.0 Methods to find the solution of first order and first degree differential equation
2.1 Variable separable method
2.2 Equations reducible to variable separable form
2.3 Homogeneous differential equation
2.4 Equations reducible to homogeneous form
2.5 Exact differential equation
2.6 Equations reducible to exact form
2.7 Linear differential equation
2.8 Equations reducible to linear form
3.0 Differential equation of first order and higher degrees
4.0 Orthogonal trajectory
2.5 Exact differential equation
2.2 Equations reducible to variable separable form
2.3 Homogeneous differential equation
2.4 Equations reducible to homogeneous form
2.5 Exact differential equation
2.6 Equations reducible to exact form
2.7 Linear differential equation
2.8 Equations reducible to linear form
A differential equation of the form $$M(x,y)dx + N(x,y)dy = 0$$ is said to be a exact differential equation if it is obtained by differentiating any function of $x$ and $y$ represented by the equation $f(x,y)=c$.
The necessary condition for the differential equation to be exact is $$\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}$$
The solution of exact differential equation is $$\int\limits_{y - {\text{constant}}} {Mdx} + \int {({\text{terms of }}N{\text{ not containing }}x)dy} = C$$
Question 9. Solve $$(2x\log y)dx + \left( {\frac{{{x^2}}}{y} + 3{y^2}} \right)dy = 0$$
Solution: Compare it with the general form of differential equation $M(x,y)dx + N(x,y)dy = 0$, we get $M(x,y) = 2x\log y$ and $N(x,y) = \frac{{{x^2}}}{y} + 3{y^2}$
$$\because \frac{{\partial M}}{{\partial y}} = \frac{{2x}}{y}{\text{ and }}\frac{{\partial N}}{{\partial x}} = \frac{{2x}}{y}{\text{ }}$$ the given differential equation is exact differential equation. The solution is $$\begin{equation} \begin{aligned} \int\limits_{y - {\text{constant}}} {(2x\log y)dx} + \int {3{y^2}dy} = C \\ \Rightarrow {x^2}\log y + {y^3} = C \\\end{aligned} \end{equation} $$