Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Reduce Order | Find Second Solution for $$y''+4y=0$$
Reply to thread
Message
[QUOTE="QuantumCurt, post: 4891823, member: 478308"] [h2]Homework Statement [/h2] [B]Given - $$y_1(x)=sin(2x)$$, find a second linearly independent solution to $$y''+4y=0$$[/B][h2]Homework Equations[/h2] [B]I'm using reduction of order to write a second solution as a multiple of the first solution. I think I've gotten to the right answer, my main question is in how I should write the general solution of the equation.[/B][h2]The Attempt at a Solution[/h2] Using the fact that ##y_1(x)=sin(2x)##, and that ##y_2(x)=vy_1(x)## I'm writing a second solution as $$y_2(x)=vsin(2x)$$ Now taking the first and second derivatives I get $$y_2'(x)=v'sin(2x)+2vcos(2x)$$ and $$y_2''(x)=v''sin(2x)+4v'cos(2x)-4vsin(2x)$$ Now I substitute back into the original equation and get, $$v''sin(2x)+4v'cos(2x)=0$$ Now to reduce order; ##z=v' and z'=v''## Then substituting - $$z'sin(2x)+4zcos(2x)=0$$ Which leads to $$\frac{dz}{z}=-4cot(2x)dx$$ Integrating this to logarithms and then exponentiating etc. leads me to - $$z=csc^2(2x)$$ Since ##z=v'## I can substitute again $$v=\int csc^2(2x)dx$$ $$v=-\frac{1}{2}cot(2x)$$ Now since ##y_2(x)=vy_1(x)##, I can say that, $$y_2(x)=[-\frac{1}{2}cot(2x)][sin(2x)]$$ which simplifies to $$y_2(x)=-\frac{1}{2}cos(2x)$$ The formula for the general solution is ##y(x)=c_1y_1(x)+c_2y_2(x)##. Given this fact, my general solution is written as - $$y(x)=c_1sin(2x)-\frac{1}{2}c_2cos(2x)$$ Now...my question - Since ##-\frac{1}{2}c_1## is really just an arbitrary constant, can I write my general solution in terms of just a constant rather than a fraction times a constant? This would give me the solution of - $$y(x)=c_1sin(2x)+c_2cos(2x)$$ Is this essentially the same solution, or would it be better to write it in terms of the negative fractional constant? Any help would be much appreciated. :) [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Reduce Order | Find Second Solution for $$y''+4y=0$$
Back
Top