- #1
roam
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Hello,
I was looking at the differential equation [tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex].
This solution to this equation has to be [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]
So, when do we usually need to use the method of integrating factor? Can we solve all linear equations (1st order) using this method?
Because in this example I used the method of separation of variables and I arrived at the correct answer!
[tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex]
[tex]\frac{dy}{dx} + xy = \frac{1}{(x^2 + 1)} [/tex]
[tex]\frac{1}{y} dy = \frac{1}{(x^2 + 1)} -xy dx[/tex]
[tex]\frac{1}{y} = \frac{-x}{(x^2 + 1)}[/tex]
[tex]\int \frac{1}{y} = \int \frac{-x}{(x^2 + 1)}[/tex]
[tex] y = - \frac{1}{2} ln (x^2 + 1) + ln c[/tex] = [tex] ln \frac{c}{\sqrt{x^2 +1}}[/tex]
& it follows that [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]!
But how do we get the same answer using the method of integrationg factor?
[tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex].
Since it is in the form [tex]P(x) \frac{dy}{dx} + Q(x)y = R(x)[/tex], we get it in the form [tex]\frac{dy}{dx} + p(x)y = q(x)[/tex];
[tex]\frac{dy}{dx} + xy = \frac {1}{x^2 +1)}[/tex];
[tex]\mu = e^{\int p(x) dx}[/tex] so, since the p(x) is just x, then ∫x = 1/2x^2
yields => [tex]e^{\frac{x^2}{2}}[/tex]
Multiplying both sides of the equation by the integrating factor;
[tex]\frac{d}{dx} + e^{\frac{x^2}{2}}y = \frac {1}{x^2 +1)} . e^{\frac{x^2}{2}}[/tex]
Did I do it correctly so far? How do we do this question using an integrating factor?
Thanks.
I was looking at the differential equation [tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex].
This solution to this equation has to be [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]
So, when do we usually need to use the method of integrating factor? Can we solve all linear equations (1st order) using this method?
Because in this example I used the method of separation of variables and I arrived at the correct answer!
[tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex]
[tex]\frac{dy}{dx} + xy = \frac{1}{(x^2 + 1)} [/tex]
[tex]\frac{1}{y} dy = \frac{1}{(x^2 + 1)} -xy dx[/tex]
[tex]\frac{1}{y} = \frac{-x}{(x^2 + 1)}[/tex]
[tex]\int \frac{1}{y} = \int \frac{-x}{(x^2 + 1)}[/tex]
[tex] y = - \frac{1}{2} ln (x^2 + 1) + ln c[/tex] = [tex] ln \frac{c}{\sqrt{x^2 +1}}[/tex]
& it follows that [tex]y = \frac{c}{\sqrt{x^2 +1}}[/tex]!
But how do we get the same answer using the method of integrationg factor?
[tex](x^2 + 1) \frac{dy}{dx} + xy = 0[/tex].
Since it is in the form [tex]P(x) \frac{dy}{dx} + Q(x)y = R(x)[/tex], we get it in the form [tex]\frac{dy}{dx} + p(x)y = q(x)[/tex];
[tex]\frac{dy}{dx} + xy = \frac {1}{x^2 +1)}[/tex];
[tex]\mu = e^{\int p(x) dx}[/tex] so, since the p(x) is just x, then ∫x = 1/2x^2
yields => [tex]e^{\frac{x^2}{2}}[/tex]
Multiplying both sides of the equation by the integrating factor;
[tex]\frac{d}{dx} + e^{\frac{x^2}{2}}y = \frac {1}{x^2 +1)} . e^{\frac{x^2}{2}}[/tex]
Did I do it correctly so far? How do we do this question using an integrating factor?
Thanks.
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