Solve the following differential equation by finding an appropriate integrating factor

Question

3xdx+(3x^(2)y+y)dy=0 The answer should be in the form F(x,y)=constant

Answer 1

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Answer 2

To solve the differential equation 3xdx+(3x2y+2y)dy=0 using an integrating factor, we first check if it's exact.

The equation is in the form M(x,y)dx+N(x,y)dy=0, where:

M(x,y)=3x

N(x,y)=3x2y+2y

For an equation to be exact, we must have ∂y∂M​=∂x∂N​.

∂y∂M​=∂y∂​(3x)=0

∂x∂N​=∂x∂​(3x2y+2y)=6xy

Since ∂y∂M​=∂x∂N​, the equation is not exact.

Now, we need to find an integrating factor. Let's try to find an integrating factor that is a function of x only, or y only.

Consider the expression N∂y∂M​−∂x∂N​​:

3x2y+2y0−6xy​=y(3x2+2)−6xy​=3x2+2−6x​

Since this is a function of x only, an integrating factor μ(x) exists, given by:

μ(x)=e∫3x2+2−6x​dx

Let u=3x2+2, then du=6xdx. So, −6xdx=−du.

∫3x2+2−6x​dx=∫u−du​=−ln∣u∣=−ln∣3x2+2∣=ln∣(3x2+2)−1∣

So, μ(x)=eln∣(3x2+2)−1∣=3x2+21​

Now, multiply the original differential equation by the integrating factor 3x2+21​:

3x2+23x​dx+3x2+23x2y+2y​dy=0

3x2+23x​dx+3x2+2y(3x2+2)​dy=0

3x2+23x​dx+ydy=0

Let's verify if this new equation is exact.

M′(x,y)=3x2+23x​

N′(x,y)=y

∂y∂M′​=∂y∂​(3x2+23x​)=0

∂x∂N′​=∂x∂​(y)=0

Since ∂y∂M′​=∂x∂N′​, the equation is now exact.

Now, we need to find a function F(x,y) such that ∂x∂F​=M′(x,y) and ∂y∂F​=N′(x,y).

Integrate M′(x,y) with respect to x:

F(x,y)=∫3x2+23x​dx+h(y)

Let u=3x2+2, then du=6xdx, so 3xdx=21​du.

F(x,y)=∫2u1​du+h(y)=21​ln∣u∣+h(y)=21​ln(3x2+2)+h(y) (since 3x2+2>0)

Now, differentiate F(x,y) with respect to y and set it equal to N′(x,y):

∂y∂F​=∂y∂​(21​ln(3x2+2)+h(y))=0+h′(y)=h′(y)

We know that ∂y∂F​=N′(x,y)=y.

So, h′(y)=y.

Integrate h′(y) with respect to y to find h(y):

h(y)=∫ydy=21​y2+C1​

Substitute h(y) back into the expression for F(x,y):

F(x,y)=21​ln(3x2+2)+21​y2+C1​

The general solution to the differential equation is F(x,y)=C, where C is a constant.

21​ln(3x2+2)+21​y2+C1​=C

21​ln(3x2+2)+21​y2=C−C1​

Since C−C1​ is just another constant, let's call it K.

21​ln(3x2+2)+21​y2=K

We can also multiply the entire equation by 2 to clear the fractions:

ln(3x2+2)+y2=2K

Let Cfinal​=2K..

The answer should be in the form F(x,y)=constant.

Final Answer: The solution to the differential equation is 21​ln(3x2+2)+21​y2=C.

Alternatively, ln(3x2+2)+y2=C.

Answer 3

3xdx + (3x²y + y)dy = 0; 3xdx + y(3x² + 1)dy = 0; ∫dy/y = - ∫3x/(3x² + 1)dx; lnIyI = - 1/2ln(3x² + 1) + lnC

IyI = C/√(3x² + 1)