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# Engineering college math and degree level student in mumbai or delhi or kolkata or bangalore

## About the lesson

Exact Differential Equation

dy/dx = f(x,y) ............(1)
A differential equation (1) of first order and of first degree can be expressed in the form
M(x,y)dx + N(x,y)dy = 0 or, simply, Mdx + Ndy = 0 .........(2)
where M and N are function of x and y .
Equation (2) is called an exact differential equation if there exists a function u(x,y) such that the term M(x,y)dx + N(x,y)dy can be expressed in the form
du = M(x,y)dx + N(x,y)dy
Then the equation becomes du = 0 and the solution is given by
u(x,y) = C
where C is an arbitrary constant.

• Maths

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### levels

• Class 10
• Higher Secondary School
• Class 12
• +4

Masters

Solve :
Here we have
{2xy cos(x2) - 2xy +1}dx + { sin(x2) - x2 + 3 }dy = 0 .........(1)

M(x,y)dx + N(x,y)dy = 0 .............(2)

comparing (1) and (2) , we get

M(x,y) = 2xy cos(x2) - 2xy +1

and N(x,y) = sin(x2) - x2 + 3

∂M/∂y = 2x cos(x2) - 2x
and
∂N/∂x = 2x cos(x2) - 2x

Therefore,
∂M/∂y = ∂N/∂x

So the given differential equation is an exact differential equation.

Therefore the solution is

⎰M(x,y) dx (y as constant) + ⎰ [term of N(x,y) is not containing x] dy = C

[ where C is an arbitrary constant]
Therefore,

⎰ {2xy cos(x2) - 2xy +1}dx + ⎰ 3 dy = C

⇒ y⎰ {2x cos(x2)}dx - 2y⎰x +⎰1dx + ⎰ 3 dy = C

[put x2 = z so that 2x = dz ]

⇒ y⎰cos(z) dz - 2y (x2/2) + x + 3y = C

⇒ y sin(z) - yx2 + x + 3y = C

⇒ y sin(x2) - yx2 + x + 3y = C

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