Question

Solve the 1st order ODE problem,
(3x+2y)dx= (3x+2y+2)dy​

Answer 1

Answer:

olve differential equation:

dydx=3x+2y3x+2y+2,y(−1)=−1

Use the following substitution:

u=3x+2y⟹dydx=12(dudx−3)

12(dudx−3)=uu+2

dudx=2uu+2+3=5u+6u+2

(u+25u+6)dudx=dx

Integrate:

∫(15+45(5u+6))dudx=∫dx

u5+425ln|5u+6|=x+C0

5u+4ln|5u+6|=25x+C1

5(3x+2y)+4ln|5(3x+2y)+6|−25x=C1

10(y−x)+4ln|15x+10y+6|=C1

5(y−x)+2ln|15x+10y+6|=C

Now we can use initial condition to find  C .

Did you perhaps forget to use absolute values for expression inside log?

C=5(−1+1)+2ln|−15–10+6|=2ln(19)

5(y−x)+2ln|15x+10y+6|=2ln(19)

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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

dx=2/dy

xy= 2