How Can I Simplify This Differential Equation to the Form dy/dx = F(ax + by)?

[chocok]

1. I need to reduce $$(a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}x +c_{2})dy = 0$$ to:

$$\frac{dy}{dx} = F ( ax + by)$$

with $$a_{1}b_{2} = a_{2}b_{1}$$

i.e.:
$$a_{1}b_{2}=a_{2}b_{1}$$

$$\frac{a_{2}}{a_{1}} = k$$ and $$\frac{b_{2}}{b_{1}} = k$$

2.
First Try: I solved for dy/dx and tired to deal with the a's and b's. But I am still left with c1 and c2.
Second Try: I tried a different approach by letting x=m+h and y=n+k, then substitute it in and try to solve for h and k. But since the coefficient of dy has 2 x's instead of a x and y, i got a system of 2 "ugly" equations that lead nowhere... (or does it?)

I tried to used the 2 methods in different ways but it seems like I am running in circle.. Can anyone pls help??

 

[gabbagabbahey]

1. I need to reduce $$(a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}x +c_{2})dy = 0$$ to:

$$\frac{dy}{dx} = F ( ax + by)$$

with $$a_{1}b_{2} = a_{2}b_{1}$$

i.e.:
$$a_{1}b_{2}=a_{2}b_{1}$$

$$\frac{a_{2}}{a_{1}} = k$$ and $$\frac{b_{2}}{b_{1}} = k$$

2.
First Try: I solved for dy/dx and tired to deal with the a's and b's. But I am still left with c1 and c2.
Second Try: I tried a different approach by letting x=m+h and y=n+k, then substitute it in and try to solve for h and k. But since the coefficient of dy has 2 x's instead of a x and y, i got a system of 2 "ugly" equations that lead nowhere... (or does it?)

I tried to used the 2 methods in different ways but it seems like I am running in circle.. Can anyone pls help??

Here's a hint: Would you consider $$\frac{u+2}{5u+10}$$ to be a function of $$u$$?

How about $$\frac{u+c_1}{ku+c_2}$$ where $$c_1,&c_2,&k$$ are constants?

What about when $$u \equiv ax+by$$ where $$a \equiv a_1$$ and $$b \equiv b_1$$?

 

[tiny-tim]

1. I need to reduce $$(a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}x +c_{2})dy = 0$$ to:

$$\frac{dy}{dx} = F ( ax + by)$$

with $$a_{1}b_{2} = a_{2}b_{1}$$
…
First Try: I solved for dy/dx and tired to deal with the a's and b's. But I am still left with c1 and c2.
… But since the coefficient of dy has 2 x's instead of a x and y, i got a system of 2 "ugly" equations that lead nowhere... (or does it?)

Hi chocok!

It must mean $$(a_{1}x + b_{1}y + c_{1})dx + (a_{2}x + b_{2}y +c_{2})dy = 0$$
…
what you've been given is obviously a misprint!

And do you need to get rid of the constants, c1 and c2? …

for example, (ax + by + c₁)² + sin(ax + by + c₂) + log(ax + by + c₃) would still be of the form F(ax + by).