Solve: dy/dx = x+y+7 / 3x+3y+5?

Question

I think it'll be a fairly easy one!!

Answer 1

this is easy because we have x+y in numerator and 3(x+y) in denominator

let x+y = t
then dy/dx + 1 = t
or dy/dx =dt/dx-1
putting in your equation

dt/dx - 1 = (t+7)/(3t+5)
or dt/dx = 1 + (t+7)/(3t+5) = (4t+12)/(3t+5)
or dx/dt = (3t+5)/4(t+3) = 1/4(3-4/(t+3))
= 3/4-1/(t+3)
integrate both sides x = 3/4t- ln(t+3) +c where c is arbitrary constant

put t = x+y we get

x= 3(x+y)/4 - ln(x+y+3) + c

Answer 2

This is a very nice problem and not nearly as easy as it looks. Let’s rewrite it carefully and see what happens.

(1) dy/dx=(x+y+7)/(3x+3y+5)

Now let Y=x+y. Substituting into (1) gives us

(2) dY/dx=1+(Y+y)/(3Y+5)=
(3) dY/dx=4(3+Y)/(3y+5)

which is separable

(4) (3y+5)/[4(3+Y)] dY=dx

Now do partial fractions

(5) (4/3+16/[5(3y+5)]) dY=dx

integrate

(6) 4Y/3+16/9 ln |3Y+5|=x+C

and resubsitute

(7) 4(x+y)+16/9 ln|3(x+y)+5|=x+C

Be sure to check the calculations carefully!

Also, do not multiply through and integrate through the sum as some have indicated. Your teacher would be very annoyed if you did so!

Good luck! :)

Answer 3

Hi Dear;

▪Step 1;

if dy/dx = x+y+7 / 3x+3y+5 ;
dx * ( x+y+7 ) = dy * ( 3x+3y+5 )

▪Step 2;
-take Integral of both sides;
∫dx ( x+y+7 ) = ∫dy ( 3x+3y+5 )
{ as you know;
∫ax + bx + c = ∫ax + ∫bx + ∫c
&
∫x^n = x ^ (n+1)/ (n+1)
& if "a" is a real number so ;∫a = 0 ... }

∫x dx + ∫y dx + ∫7 dx = ∫3x dy +∫3y dy + ∫5 dy
[ x^2/2 + xy ] = [ 3xy + (3/2)y^2]
x^2/2 + xy - 3xy - (3/2)y^2 = 0
x^2/2 -2xy - (3/2)y^2 = 0
{ mulityply this by ' 2' }
2( x^2/2) - 2(2xy) - 2(3/2)y^2 = 0
x^2 -4x - 3y^2 = 0

Good Luck

Answer 4

dy / dx = x + y + 7 / 3x + 3y + 5

=> dy(3x + 3y + 5) = dx(x + y + 7)
integrating it,
=> ∫ dy × 3x + ∫ 3y dy + ∫ 5 = ∫ x dx + ∫ y dx + ∫ 7
=> 3xy + 3/2 y² = x²/2 + xy
=> 2xy + 3/2 y² - x²/2 = 0
=> x² - 4xy - 3y² = 0

Answer 5

What exactly are you supposed to be solving, since you already have dy/dx? Are you supposed to find vertical and horizontal tangents?

Horizontal tangents: numerator must be zero, and denominator must be nonzero. In this case, you must have x+y=-7, so you'd need to substitute y=-7-x into the equation for the original curve to find the points at which this occurs. If x+y+7=0, then 3x+3y+5=-16, so you don't have to worry about getting 0/0.

Vertical tangents: other way around. See when 3x+3y+5=0. Solve this for y and substitute into the equation for your original curve to find the points at which there is a vertical tangent.

If that's not what you're supposed to solve for, then please clarify your question.

Answer 6

Easy for YOU to say!