# Not sure of the method to use

#### david.a491

##### New member
Hello everyone
I'm trying to solve the following equation with the easiest approach possible

(3x+2y)dx+(3x+2y-5)dy=0

I did some work to it

(3x+2y)dx+(3x+2y)dy-5dy=0

(3x+2y)(dx+dy)=5dy

The right side is ready to integrate, however the constants on the left side are completely throwing me off, if I had (x+y)(dx+dy) I'd be ready to integrate, but the 3 and the 2 are in my way and I see no way to get rid of them, any help would be appreciated!

#### HallsofIvy

##### Elite Member
Seeing that "3x+ 2y" in two different places I would try letting u= 3x+ 2y. Then du= 3dx+ 2dy so that 2dy= du- 3dx and dy= (1/2)du- (3/2)dx. The differential equation becomes udx+ (u- 5)((1/2)du+ (3/2)dx)= (u- 3/2)dx+ (1/2)(u- 5)du= 0.

$$\displaystyle dx= -\frac{1}{2}\frac{u-5}{u- \frac{3}{2}}du$$.

• • david.a491 and Subhotosh Khan

#### david.a491

##### New member
Seeing that "3x+ 2y" in two different places I would try letting u= 3x+ 2y. Then du= 3dx+ 2dy so that 2dy= du- 3dx and dy= (1/2)du- (3/2)dx. The differential equation becomes udx+ (u- 5)((1/2)du+ (3/2)dx)= (u- 3/2)dx+ (1/2)(u- 5)du= 0.

$$\displaystyle dx= -\frac{1}{2}\frac{u-5}{u- \frac{3}{2}}du$$.

Thank you so much! From there I'm ready to integrate, long division and it becomes really simple after! What a life saver