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/sci/ - Science & Math

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1207188 No.1207188 [Reply] [Original]

Hi! Can anyone please check this out for me?

Solve the following obtaining an expression for <span class="math">y^2[/spoiler] in terms of <span class="math">x[/spoiler].

<span class="math">y*\frac{dy}{dx}=1+y^2[/spoiler]
<span class="math">y=2\,x=0[/spoiler]

<span class="math">\int {\frac{y}{1+y^2}}\,dy=\int\,dx[/spoiler]
<span class="math">ln|1+y^2|=x+\frac{ln|5|}{2}[/spoiler]
<span class="math">y^2={\frac{5e^x}{e^2}}-1[/spoiler]

Thanks!

>> No.1207204
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1207204

Solve it yourself you lazy ass

>> No.1207212

>>1207188

Anyone?

I really need a bit of help here?

I'm trying to learn calculus on my own and I have no reference for problems other than you /sci/ducks!

>>1207204
It's solved. I just want someone to tell me if it's right or not?!

>> No.1207213

>exp(ln(5)/2)=5/exp(2)
This looks legit.

>> No.1207214

>>1207188
you forgot constant
faggot

>> No.1207236

<div class="math">y*\frac{dy}{dx}=1+y^2</div>
>ok
<div class="math">y=2\,x=0</div>
>wat. (let's just skip this)
<div class="math">\int {\frac{y}{1+y^2}}\,dy=\int\,dx</div>
>correct
<div class="math">ln|1+y^2|=x+\frac{ln|5|}{2}</div>
>this is where the bullshit starts
<div class="math">\int {\frac{y}{1+y^2}}\,dy = \frac{1}{2}\ln{(1 + y^2)} + C</div>

>> No.1207246

>>1207188

Kk Op here.

<span class="math">\int {\frac{y}{1+y^2}}\,dy=\int\,dx[/spoiler]
<span class="math">\frac{ln|1+y^2|}{2}=x+\frac{ln|5|}{2}[/spoiler]
<span class="math">y^2=5e^{2x} -1[/spoiler]

Better?

>> No.1207250

>>1207236
continued
<div class="math">\frac{1}{2}\ln{(1 + y^2)} = x + C</div>
<div class="math">1 + y^2 = Ce^{2x}</div>
<div class="math">y = \pm \sqrt{Ce^{2x} - 1}</div>

>> No.1207252

>>1207214

>constant

Is already worked out idiot.

>>1207236

Yeah that was an omission in my working. I corrected it?!

I still don't know if the overall method is correct?!

>> No.1207257

>>1207246
correct now

>> No.1207264

>>1207236
>>1207250
>>1207257

Thanks for helping out :)