Homogeneous Differential Equation: how to do this equation ???

Question

x^2+y'=4x^2+7xy+2y^2

 

i let: y=ax ; dy= adx + xda

then substitute,

 

x^2 (adx + xda) = 4x^2 + 7x(ax) +2(ax)^2

adx + xda = 4 + 7a + 2a^2

∫dx/x + ∫da/a + (2a-7)/x = 4/ax

ln x + ln a + ???????????

cant answer next procedure

Comments

Please check that you have stated the DE correctly. I don’t think there should be a plus sign on the left of the equals, and your own workings out seem to confirm this.

Answer 1

I assume there is an error in your DE, otherwise it would not be homogeneous. It should be:

x²y'=4x²+7xy+y².

Using your substitution y=ax, y'=a+xda/dx.

The DE becomes:

x²(a+xda/dx)=4x²+7ax²+2a²x²=x²(4+7a+2a²).

Assuming x≠0,

a+xda/dx=4+7a+2a², xda/dx=4+6a+2a²=2(2+3a+a²)=2(1+a)(2+a).

Now the variables can be separated:

da/(2(1+a)(2+a))=dx/x.

This can be expressed as partial fractions:

1/(2(1+a)(2+a))=½(A/(1+a)+B/(2+a)), where A and B are constants.

So A(2+a)+B(1+a)=1, 2A+B=1 (constant), A+B=0 (coefficient of x term), therefore B=-A, making 2A-A=1, that is, A=1, and B=-1. So we now have ½(da/(1+a)-da/(2+a))=dx/x.

Now we can integrate: ½ln|1+a|-½ln|2+a|=ln|cx| where c is a constant.

We can write this: ln|(1+a)/(2+a)|=ln(cx)², and equating the logs:

(1+a)/(2+a)=c²x².

Now put a=y/x:

(x+y)/(2x+y)=kx² where constant k=c².

The solution can be expressed in another form:

x+y=kx²(2x+y)=2kx³+kx²y,

y(1-kx²)=x(2kx²-1), y=x(2kx²-1)/(1-kx²).

QUICK CHECK

Let k=0, then y=-x is the solution. y'=-1 and the DE becomes:

-x²=4x²-7x²+2x², which is true.