Solve for the following problems using substitution suggested by the

William Collins

William Collins

Answered question

2021-12-19

Solve for the following problems using substitution suggested by the equation. (3x2y+1)dx+(3x2y+3)dy=0

Answer & Explanation

Anzante2m

Anzante2m

Beginner2021-12-20Added 34 answers

Step 1
We have the given differential equation as
(3x2y+1)dx+(3x2y+3)dy=0
On solving further, we get the result as
(3x2y+3)dy=(3x2y+1)dx
dydx=(3x2y+1)(3x2y+3)
Now, let us consider that
3x2y=u
32dydx=dudx
3dudx=2dydx
dydx=12(3dudx)
Step 2
Therefore, 12(3dudx)=(u+1)u+3
3dudx=2u+2u+3
dudx=3+2u+2u+3
dudx=3u+9+2u+2u+3
dudx=5u+11u+3
(u+35u+11)du=dx
On integrating both of the side, we get the result as
u+35u+11du=dx
155u+11+45u+11du=x+c
Foreckije

Foreckije

Beginner2021-12-21Added 32 answers

Given differential equation is
(3x2y+1)dx+(3x2y+3)dy=0
Put 3x2y=u,then 32dydx=dudx
2dydx=dudx3
dydx=12dudx32
Given differential equation can be written as
dydx=3x2y+13x2y+3
12dudx32=u+1u+3
12dudx=u+1u+3+32
dudx=2u+2u+3+3
dudx=2u2+3u+9u+3
dudx=u+7u+3
u+3u+7du=dx
(14u+7)du=dx
(14u+7)du=dx+c
u4ln(u+7)=x+c
3x2y4ln(3x2y+7)=x+c
RizerMix

RizerMix

Expert2021-12-29Added 656 answers

(3x2y+1)2x+(3x2y+3)dy=0dydx=(3x2y+1)(3x2y+3)3x2y=u32dydx=dudxdydx=12(3dudx)=3212dudxNow,3212dudx=(4+1)(4+3)32+4+14+3=12dudx34+9+24+22(4+3)=12dudx54+112(4+3)=12dudx54+11u+3=dudxdx=u+354+11dudx=(15+45(54+11))dx=15u+425ln(54+11)+cx=3x2y5+425(5(3x2y)+11)+c2x2y5+425ln(5(3x2y)+11)+c=0

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?