Exact Differential Equations

An equation that involves a differential co-efficient is called a Differential equation. A differential equation of the form that includes one or more terms as well as the derivatives of one variable (the dependent variable) with respect to another variable (i.e., independent variable)

dy/dx = f(x)

where, “x” is an independent variable, while “y” is a dependent variable in this case.

Example:  dy/dx = 2x

Ordinary differential equation

A Differential equation is said to be an ordinary differential equation if the derivatives in the equation are ordinary derivatives.

Example:

1. dy/dx + 12xy = x²
2. d²y/dx² + 12dy/dx + 9y = e^x

Partial differential equation

A differential equation is said to be a partial differential equation if the derivatives in the equation have reference to two or more independent variables (i.e. partial derivatives are used in the given equation).

Example: ∂²u/∂x²=12∂²u/∂t² ( Where t & x are independent variables )

Order of the differential equation

The order of the differential equation is the order of the equation’s highest order derivative.

First Order Differential Equation

As seen in the example below, it is a first-order differential equation with a degree equal to 1. All linear equations in the form of derivatives are of the first order. It only has the first derivative, dy/dx, where x and y are the two variables, and is denoted as

dy/dx = f(x,y) = y^‘

Example: dy/dx = 2x+3

Second-Order Differential Equation

The equation that contains the second-order derivative is known as the second-order differential equation. It is written as follows:

d/dx(dy/dx) = d²y/dx² = f^”(x) = y^”

Degree of the differential equation

The power of the highest derivative in the differential equation is called the degree of the differential equation. The degree of the differential equation is equal to the power of the highest order derivative, where the original problem is represented as a polynomial equation with derivatives such as y’, y”, y”‘, etc.

Example:

(d²y/dx²)⁴ + (d³y/dx³)² + cos x = 0

Here order= 3 and degree = 2

(y^”’)³ + 12y^” + 8y^‘ + 16 = 0

Here degree is 3

Homogeneous function

A function f(x, y) is said to be homogeneous function if f( kx, ky ) = kⁿf( x, y) where ‘n’ is the degree

Example:

f(x,y) = (x³ + y³)/(x+y)

f(kx + ky)= [(kx)³ + (ky)³]/ (kx + ky)

               = k³(x³ + y³)/k(x + y)

               = k²f(x,y)

It is a degree 2 homogeneous function.

First-order and first-degree Differential equation

An equation of the form dy/dx = f(x, y) is said to be first order and first-degree differential equation.

Homogeneous Differential equation

A differential equation dy/dx = f(x, y) is said to be a homogeneous differential equation if f(x, y) is a homogeneous function and the degree must be zero

Example:

 let f(x,y) = (x² + y²)/2xy

f(kx,ky) = [(kx)²+(ky)²]/2(kx)(ky)

            = k²(x² + y²)/k²(2xy)

            = k⁰ f(x,y)

therefore f(x, y) is homogeneous function and degree is 0

Exact differential equations

A differential equation Mdx + Ndy = 0 is said to be exact differential equation, if ∂M/∂y =  ∂N/∂x

Working rule to solve exact differential equations

Step 1: Given differential equation can be written as Mdx + Ndy = 0 form considering as equation 1

Step 2: Check ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is

∫Mdx +∫Ndy =C      

(y= constant) and (do not contain x)

Solving exact equations by Integrating factor

If Mdx + Ndy = 0 is a homogeneous differential equation and Mdx + Ndy is not equal to 0 the 1/(Mx+Ny) is a integrating factor of Mdx + Ndy = 0.

Example: Solve the exact differential equation x²ydx−(x³+y³)dy=0

Solution:

Comparing with Mdx + Ndy = 0

M = x²y, and N = -(x³+y³)

∂M/∂y = x²       and ∂N/∂x = -3x²

So here ∂M/∂y is not equal to ∂N/∂x

Here the given equation is not exact differential equation

Considering Integrating factor

Integrating Factor =1/Mx+Ny

=1/(x²y)x+(−x³−y³)y

=  1/(x³y-x³y-y⁴)=-1/y⁴

=−1/y⁴ is an integrating factor

Multiplying the equation with Integrating factor

=(−x²y/y⁴)dx+(x³+y³/y⁴)dy=0

=(−x²/y³)dx+(x³/y⁴+1/y)dy=0

Comparing with  M₁dx + N₁dy= 0

M₁= -x²/y²    and N₁ = x³/y⁴+1/y

Therefore ∂M₁/∂y =  ∂N₁/∂x

Finding ∫M₁dx +∫N₁dy =C      

∫(−x²/y³)dx+(x³/y⁴+1/y)dy=C

−x³/3y³+logy=C

It is the solution for given equation.

Sample Problems

Problem 1: Solve (hx + by + f)dy + (ax + hy + g)dx = 0

Solution:

Given equation is (hx+by+f)dy+(ax+hy+g)dx=0….(1)

Step 1: Comparing with Mdx + Ndy = 0

M= ax+hy+g and N = hx+by+f

Step 2: Check ∂M/∂y =  ∂N/∂x

 ∂M/∂y = h

 ∂N/∂x = h

Therefore  ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is 

∫Mdx +∫Ndy =C

(y= constant) and (do not contain x)

=∫(ax+hy+g)dy+∫(hx+by+f)dx=C

=ax²/2+hy∫dx+g∫dx+0+by²/2+f∫dy=C

=ax²/2+hyx+gx+by²/2+fy=C  ( here ∫dx= x, ∫dy = y)

Therefore it is the solution for given differential equation.

Problem 2: Solve (y²−2xy)dx−(x²−2xy)dy=0

Solution:

Given equation is (y²−2xy)dx−(x²−2xy)dy=0….(1)

Step 1: Comparing with Mdx + Ndy = 0

M= y²-2xy and N = -x²+2xy

Step 2: Check ∂M/∂y =  ∂N/∂x

∂M/∂y = 2y-2x

∂N/∂x = -2x+2y

Therefore  ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is

∫Mdx +∫Ndy =C

(y= constant) and (do not contain x)

=∫(y²−2xy)dx+∫(−x²+2xy)dy=C

=y²∫dx−2y∫xdx+0=C

=xy²−x²y=C

Therefore it is a solution for a given differential equation.

Problem 3: Solve (y(1+1/x)+cosy)dx+(x+logx−siny)dy=0

Solution:

Given equation is (y(1+1/x)+cosy)dx+(x+logx−siny)dy=0

Step 1: Comparing with Mdx + Ndy = 0

M= y+y/x+cosy and N = x+logx-xsiny

Step 2: Check ∂M/∂y =  ∂N/∂x

∂M/∂y = 1+1/x-siny

∂N/∂x = 1+1/x-siny

Therefore  ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is

∫Mdx +∫Ndy =C

(y= constant) and (do not contain x)

∫(y+y/x)+cosy)dx+∫(x+logx−siny)dy=C

=xy+ylogx+xcosy+0=C

=y(x+logx)+xcosy=C

Therefore it is a solution for a given differential equation.

Problem 4: Solve (x²−4xy−2y²)dx+(y²−4xy−2x²)dy=0

Solution:

Given equation is (x²−4xy−2y²)dx+(y²−4xy−2x²)dy=0

Step 1: Comparing with Mdx + Ndy = 0

M= x²-4xy-2y² and N = y²-4xy-2x²

Step 2: Check ∂M/∂y =  ∂N/∂x

∂M/∂y = -4x-4y

∂N/∂x = -4x-4y

Therefore  ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is

Mdx +∫Ndy =C

(y= constant) and (do not contain x)

=∫(x²−4xy−2y²)dx+∫(y²−4xy−2x²)dy=C

=x³/3−4yx²/2−2y²x+y³/3=C

=x³/3−2x²y−2xy²+y³/3=C

Therefore it is a solution for given differential equation.

Problem 5: Solve (2xy+y-tany)dx+(x²-xtan²y+sec²y)dy=0

Solution:

Given equation is (2xy+y-tany)dx+(x²-xtan²y+sec²y)dy=0

Step 1: Comparing with Mdx + Ndy = 0

M= 2xy+y-tany and N = x²-xtan²y

Step 2: Check ∂M/∂y =  ∂N/∂x

∂M/∂y = 2x+1-sec²y = 2x-tan²y

∂N/∂x = 2x-tan²y

Therefore  ∂M/∂y =  ∂N/∂x

Step 3: The general solution of equation 1 is

∫Mdx +∫Ndy =C

(y= constant) and (do not contain x)

=∫(2xy+y−tany)dx+∫(x²−xtan²y+sec²y)dy=C

=2y∫xdx+y∫dx−tany∫dx+0+0+∫sec²ydy=C

=2yx²/2+xy−xtany+tany=C

=xy(1+x)+tany(1−x)=C

Therefore it is a solution for a given differential equation.