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College Discussion / SAT and ACT Tests & Test Preparation / AP Tests Preparation / Mathematics & Computer Science

PettyOfficer
Registered User Posts: **77** Junior Member

Q: Solve the differential equation :

(x+y+1)(dy/dx) = 1

Ans : I put (x+y)=t

=> y=t-x

(t+1)(d(t-x)/dx) = 1

(t+1)(dt/dx - 1) = 1

(t+1)(dt/dx) = t+2

[(t+1)/(t+2)]*(dt/dx) = dx

integrating both sides,

int[ t/(t+2) + 1/(t+2) ]dt = int.dx

now the second term in the LHS can be integrated directly, and so can the RHS.

but the term t/(t+2) has to be integrated separately by adding and subtracting 2 in numerator.

=> (t+2-2)/(t+2)

=> 1 - 2/(t+2)

and so this can now be solved too.

=> int.[ 1-2/(t+2) + 1/(t+2) ]dt = int.dx

=> int.[ 1 - 1/(t+2) ]dt = int.dx

=> t - log(t+2) = x +C

substituting t= x+y

x+y - log(x+y+2) = x +C

y= log(x+y+2) +C

which i think is the answer.

am I correct? I dont have the answers so if some1 can verify, It would be a great.

(x+y+1)(dy/dx) = 1

Ans : I put (x+y)=t

=> y=t-x

(t+1)(d(t-x)/dx) = 1

(t+1)(dt/dx - 1) = 1

(t+1)(dt/dx) = t+2

[(t+1)/(t+2)]*(dt/dx) = dx

integrating both sides,

int[ t/(t+2) + 1/(t+2) ]dt = int.dx

now the second term in the LHS can be integrated directly, and so can the RHS.

but the term t/(t+2) has to be integrated separately by adding and subtracting 2 in numerator.

=> (t+2-2)/(t+2)

=> 1 - 2/(t+2)

and so this can now be solved too.

=> int.[ 1-2/(t+2) + 1/(t+2) ]dt = int.dx

=> int.[ 1 - 1/(t+2) ]dt = int.dx

=> t - log(t+2) = x +C

substituting t= x+y

x+y - log(x+y+2) = x +C

y= log(x+y+2) +C

which i think is the answer.

am I correct? I dont have the answers so if some1 can verify, It would be a great.

Post edited by PettyOfficer on

This discussion has been closed.

## Replies to: Is The solution of differential equation correct?

4,747Senior MemberA slightly shorter way to integrate the LHS is to see that (t+1)/(t+2) = (t+2)/(t+2) - 1/(t+2).

77Junior Memberand i didn't see that shortcut, nice thinking.