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## Main Question or Discussion Point

Hello!

Here's a question I was doing but I'm not sure if I'm right...

It says "solve" the following differential equation:

[tex](x^2 +1) \frac{dy}{dx} = 0[/tex]

So, this is what I've done:

We put it in form [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

Then we devide it by x^2 +1 to obtain:

[tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

you know , in standard form where p(x) = x/x^2 +1

Integrating factor is:

[tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]

[tex]= exp[ln(x^2 +1)^{2}][/tex]

[tex]= (x^2 +1)^{2}[/tex]

Multiplying the DE by the integrating factor;

[tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

Now we integrate:

[tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]

=> [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

Am I right? Please correct me if I'm wrong...

Thank you,

Here's a question I was doing but I'm not sure if I'm right...

It says "solve" the following differential equation:

[tex](x^2 +1) \frac{dy}{dx} = 0[/tex]

So, this is what I've done:

We put it in form [tex]\frac{dy}{dx} +p(x)y = q(x)[/tex]

Then we devide it by x^2 +1 to obtain:

[tex]\frac{dy}{dx} + \frac{x}{x^2 +1} y = \frac{1}{x^2 +1} \times 0[/tex]

you know , in standard form where p(x) = x/x^2 +1

Integrating factor is:

[tex]exp[\int p(x) dx = exp[\frac{x dx}{x^2 +1}[/tex]

[tex]= exp[ln(x^2 +1)^{2}][/tex]

[tex]= (x^2 +1)^{2}[/tex]

Multiplying the DE by the integrating factor;

[tex](x^2 +1)^{2} \frac{dy}{dx} +x(x^2 +1)y = x . (x^2 +1)^{2}[/tex]

Now we integrate:

[tex]\frac{d}{dx} [(x^2 +1)^{2} y] = x^3 +x[/tex]

=> [tex][(x^2 +1)^{2} y] = \frac{x^4}{4} + \frac{x^2}{2} + c [/tex]

Am I right? Please correct me if I'm wrong...

Thank you,