# Thread: 2 more questions

1. ## 2 more questions

once the integral is found for dy/dx=y^2+1, how do you find the equation that satisfies the initial condition of y(4)=0?

i put y=tan(x+4), but it's wrong.

2. Originally Posted by sw3etazngyrl
once the integral is found for dy/dx=y^2+1, how do you find the equation that satisfies the initial condition of y(4)=0?

i put y=tan(x+4), but it's wrong.
i got y = tan(x - 4). what was the answer?

3. i don't know, it doesn't tell me. it only tells me that my answer is wrong.

4. how did you get that answer?

5. Originally Posted by sw3etazngyrl
i don't know, it doesn't tell me. it only tells me that my answer is wrong.
oh. you're inputing answers into some online application? is y = tan(x - 4) also wrong?

we have $y' = y^2 + 1$

$\Rightarrow \frac {y'}{y^2 + 1} = 1$

$\Rightarrow \arctan y = x + C$

$\Rightarrow y = \tan(x + C)$

Now $y(4) = 0$

$\Rightarrow 0 = \tan (4 + C)$

$\Rightarrow \arctan0 = 4 + C$

$\Rightarrow 0 = 4 + C$

$\Rightarrow C = -4$

Thus our solution is $y = \tan (x - 4)$

6. the answer you got is right. thanks! but how did you arrive at that?

7. $\frac{{dy}}
{{dx}} = y^2 + 1 \implies \frac{1}
{{y^2 + 1}}\,dy = dx \implies \arctan y = x + k$

It's just arctan formula.