# Check My Work for solving this Differential Equation? :)

1. Feb 3, 2016

### Dusty912

1. The problem statement, all variables and given/known data
Question: Solve this differential equation
dy/dt=1+y2 and y(π/4)=-1
2. Relevant equations

3. 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

2. Feb 3, 2016

### Samy_A

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

3. Feb 3, 2016

### Dusty912

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

4. Feb 3, 2016

### Samy_A

Yes.

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

5. Feb 3, 2016

### Dusty912

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

6. Feb 3, 2016

Yes.

7. Feb 3, 2016

### Dusty912

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

8. Feb 3, 2016

### Samy_A

Yes (no need for guessing, though).

9. Feb 3, 2016

### 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)

10. Feb 3, 2016

### Samy_A

Yes.

11. Feb 3, 2016

### Dusty912

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