Question

(4x³ + 3x²y²)dx + 2x³ y dy =0 solve the differential equation

Answer 1

Explanation:

How do you solve the equation (cos2y−3x2y2)dx+(cos2y−2xsin2y−2x3y)dy=0 using methods of exact differential equations?

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(cos2y−3x2y2)dx+(cos2y−2xsin2y−2x3y)dy=0

In standard form,

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

dydx=f(x,y)

We also assume that

∂F∂x=M(x,y)

∂F∂y=N(x,y)

Let

M(x,y)=(cos2y−3x2y2)

N(x,y)=(cos2y−2xsin2y−2x3y)

Let's test if the differential belongs to an 'EXACT' differential equation!

We say that a differential is an exact differential equation if the first order partial derivative of the differential exists and equal to the other side,

∂M(x,y)∂y=∂N(x,y)∂x

This is known as Clairaut's theorem,

Definition,

If the second order mixed partial derivatives which exist and are continuous on the open disk which that contains point (a,b)

∂2F∂y∂x=∂2F∂x∂y

That theorem will implies continuity!

Let's test it now,

∂M(x,y)∂y=(−2sin2y−6x2y)

∂N(x,y)∂x=(−2sin2y−6x2y)

It belongs to exact differential equation.

Let's solve it

Method I:(standard method also known as the long way)

Integration M(x,y) with respect to dx ,

F(x,y)=(xcos2y−x3y2+h(y))

Now, differentiate partially w.r.t y then,

Fy(x,y)=(−2xsin2y−2x3y+h′(y))

Comparing with

N(x,y)=(cos2y−2xsin2y−2x3y)

We can conclude that

h′(y)=cos2y

Integrating with respect to y again,

h(y)=sin2y2+c

So the original function is then,

F(x,y)=xcos2y−x3y2+sin2y2+c

Which is then xcos2y−x3y2+sin2y2=C

c&C are known as the constant of integration.

Method 2:(Grouping method)

This method is a lot quicker and taking less effort! Seeing which and which are produced by the same function, it is advisable if you know the integral of the derivative well.

(cos2y−3x2y2)dx+(cos2y−2xsin2y−2x3y)dy=(−3x2y2dx−2x3ydy)+(cos2ydx−2xsin2ydy)+(cos2y)dy

(−3x2y2dx−2x3ydy)+(cos2ydx−2xsin2ydy)+(cos2y)dy=0

Integrating with respect to each variable, we have

(−x3y2)+(xcos2y)+(sin2y2)=c

We know that whatever constant we differentiate it must be 0 !

dcdy=0

This two methods are only working for EXACT DIFFERENTIAL EQUATION!

If ∂M(x,y)∂y≠∂N(x,y)∂x, then it is not applicable.

What ever method we use we end up in integral curves of family of solution because the c is not determined!

We may try with method of integrating facto