Check My Work for solving this Differential Equation? :)

[Dusty912]

Homework Statement

Question: Solve this differential equation
dy/dt=1+y² and y(π/4)=-1

Homework Equations

The Attempt at a Solution

dy/dt=1+y² and y(π/4)=-1

dy/dt=1+y²
dy=(1+y²)dt
dy/(1+y²)=dt
∫dy(1+y²)=∫dt
tan^-1(y)=t+c
y(t)=tan(t)+c
y(π/4)=tan(π/4)+c
-1=1+c
-2=c

answer: y(t)=tan(t)-2

 

[Samy_A]

Homework Statement

Question: Solve this differential equation
dy/dt=1+y² and y(π/4)=-1

Homework Equations

The Attempt at a Solution

dy/dt=1+y² and y(π/4)=-1

dy/dt=1+y²
dy=(1+y²)dt
dy/(1+y²)=dt
∫dy(1+y²)=∫dt
tan^-1(y)=t+c
y(t)=tan(t)+c
y(π/4)=tan(π/4)+c
-1=1+c
-2=c

answer: y(t)=tan(t)-2

Did you check if dy/dt=1+y² for your solution?

I think there is an error in this step:
tan^-1(y)=t+c
y(t)=tan(t)+c

 

[Dusty912]

Did you check if dy/dt=1+y² for your solution?

I think there is an error in this step:
tan^-1(y)=t+c
y(t)=tan(t)+c

hmm I'm not sure that I am seeing the error, does it involve the constant being effected by the tan?

 

[Samy_A]

hmm I'm not sure that I am seeing the error, does it involve the constant being effected by the tan?

Yes.

There must be an error somewhere, as inspection would show that y(t)=tan(t)-2 doesn't satisfy the differential equation.

 

[Dusty912]

so would it look something like this: y=tan(t+c) ----not my answer but the next step

 

[Samy_A]

so would it look something like this: y=tan(t+c) ----not my answer but the next step

Yes.

 

[Dusty912]

ok so then then from there using y(π/4)=-1
the next step would be: -1=tan(π/4+c)
and then I am guessing the step would be: tan^-1(-1)=π/4+c

 

[Samy_A]

ok so then then from there using y(π/4)=-1
the next step would be: -1=tan(π/4+c)
and then I am guessing the step would be: tan^-1(-1)=π/4+c

Yes (no need for guessing, though).

 

[Dusty912]

adding on to the previous steps: -π/4=π/4+c
-π/2=c

so the the solution would be:
y(t)=tan(t-π/2)

 

[Samy_A]

adding on to the previous steps: -π/4=π/4+c
-π/2=c

so the the solution would be:
y(t)=tan(t-π/2)

Yes.

 

[Dusty912]

Okay thank you very much. You've been very helpful.