Differential Equation: (x^2+3xy+y^2)dx-x^2dy=0
In this article, we will discuss the differential equation (x^2+3xy+y^2)dx-x^2dy=0
and its solution.
Form of the Differential Equation
The given differential equation is of the form:
M(x,y)dx+N(x,y)dy=0
where M(x,y)=x^2+3xy+y^2
and N(x,y)=-x^2
.
Solution of the Differential Equation
To solve this differential equation, we can use an integrating factor. Let's find the integrating factor:
μ(x,y)=e^∫(N_x-M_y)/M dx
where N_x
is the partial derivative of N
with respect to x
, and M_y
is the partial derivative of M
with respect to y
.
First, let's find the partial derivatives:
N_x=-2x
M_y=3x+2y
Now, we can find the integrating factor:
μ(x,y)=e^∫(-2x-(3x+2y))/(x^2+3xy+y^2) dx
μ(x,y)=1/(x^2+3xy+y^2)
Now, multiply the original differential equation by the integrating factor:
(1/(x^2+3xy+y^2))(x^2+3xy+y^2)dx-(1/(x^2+3xy+y^2))x^2dy=0
This simplifies to:
dx-(x^2/(x^2+3xy+y^2))dy=0
Solving the Differential Equation
To solve this differential equation, we can integrate both sides with respect to x
, treating y
as a constant:
∫dx-∫(x^2/(x^2+3xy+y^2))dy=C
where C
is the constant of integration.
The first integral is simple:
x=C1
The second integral is more complicated:
-∫(x^2/(x^2+3xy+y^2))dy=C2
To evaluate this integral, we can use the substitution u=x^2+3xy+y^2
, du=(2x+3y)dx
. Then, the integral becomes:
-∫(1/u)du=C2
which evaluates to:
ln|x^2+3xy+y^2|=C2
Now, we can combine the two results:
x-C ln|x^2+3xy+y^2|=C
which is the general solution of the differential equation.
Conclusion
In this article, we have solved the differential equation (x^2+3xy+y^2)dx-x^2dy=0
using an integrating factor. The general solution of the differential equation is x-C ln|x^2+3xy+y^2|=C
, where C
is the constant of integration.