Exact Differential Equations

Equation that involves a differential co-efficient is called a Differential equation. Let's take the equation P(x, y)dx + Q(x, y)dy = 0. This equation is called the exact differential equation if, ∂P/∂y = ∂Q/∂y.

In this article, we have covered in detail about exact differential equation.

Exact Differential Equation Definition

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

Testing for Exactness

Rake functions P(x, y) and Q(x, y) having the continuous partial derivatives in a particular domain, then the differential equation is exact iff,

∂P/∂y = ∂Q/∂y

Solving Exact Differential Equations

We can solve exact differential equation by following the steps added below as:

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: General solution of equation 1 is

∫Mdx +∫Ndy =C      

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

Solving Exact Differntial Equations by Integrating factor

If Mdx + Ndy = 0 is a homogeneous differential equation and Mdx + Ndy is not equal to 0 then this equation can be solve by multiplying the differential equation with integrating factor and the integrating factor for the same is, 1/(Mx+Ny) for equation of the form Mdx + Ndy = 0.

Example: Solve 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, 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.

Exact Differential Equation Examples

Examples of exact differential equations are:

• (2xy + y³)dx + (x² + 3xy²)dy = 0
• (3x²y - y³)dx + (x³ - 3xy²)dy = 0
• (2xy + y²)dx + (x² + 2xy)dy = 0
• (x²cosy - ysinx)dx + (xsiny + ycosx)dy = 0

Read More:

• Linear Differential Equation
• Application of Differential Equation

Exact Differential Equation 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.