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Solving a first order differential equation

LLand314

Member
Mar 26, 2013
77
cosx(dy/dx) + ysinx = sinx cosx

(dy/dx) + y/cosx = 1

e^integral (1/cosx) ?

I feel like this has to do with ln again but not sure
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
In your attempt to write the equation in standard linear form, you have made a few errors when dividing through by $\cos(x)$. Try it again and see what you get. :D
 

LLand314

Member
Mar 26, 2013
77
In your attempt to write the equation in standard linear form, you have made a few errors when dividing through by $\cos(x)$. Try it again and see what you get. :D

dy/dx + ysinx/cosx = (sinx cosx)/cosx

dy/dx + ytanx = tanx ?

Or would I divide sinx first and move it over the other side before I divide cosx
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have the left side correct as \(\displaystyle \frac{\sin(x)}{\cos(x)}=\tan(x)\), but the right side would simply have \(\displaystyle \cos(x)\) cancelling, leaving \(\displaystyle \sin(x)\).

Your goal in arranging the equation in standard linear form is to get it in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

and since this equation was given in the form:

\(\displaystyle f(x)\frac{dy}{dx}+g(x)y=h(x)\)

you simply need to divide through by $f(x)$ to get in the the standard form.
 

LLand314

Member
Mar 26, 2013
77
You have the left side correct as \(\displaystyle \frac{\sin(x)}{\cos(x)}=\tan(x)\), but the right side would simply have \(\displaystyle \cos(x)\) cancelling, leaving \(\displaystyle \sin(x)\).

Your goal in arranging the equation in standard linear form is to get it in the form:

\(\displaystyle \frac{dy}{dx}+P(x)y=Q(x)\)

and since this equation was given in the form:

\(\displaystyle f(x)\frac{dy}{dx}+g(x)y=h(x)\)

you simply need to divide through by $f(x)$ to get in the the standard form.
So is my integrating factor 1/cosx ?

So it becomes (dy/dx)(1/cosx) + (ytanx)(1/cosx) = tanx

Integral (ytanx)(1/cosx) = integral tanx

y(1/cosx) = ln|secx| + C
y= cosx ln|secx| + Ccosx ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, I would write:

\(\displaystyle \mu(x)=e^{\int\tan(x)\,dx}=\sec(x)\) and so we have:

\(\displaystyle \sec(x)\frac{dy}{dx}+\sec(x)\tan(x)y=\tan(x)\)

\(\displaystyle \frac{d}{dx}\left(\sec(x)y \right)=\tan(x)\)

\(\displaystyle \int\,d\left(\sec(x)y \right)=\int\tan(x)\,dx\)

\(\displaystyle \sec(x)y=\ln|\sec(x)|+C\)

\(\displaystyle y(x)=\cos(x)\left(\ln|\sec(x)|+C \right)\)