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
Help Solving the following Differential equation
Reply to thread
Message
[QUOTE="Jose Z, post: 5463729, member: 592952"] [h2]Homework Statement [/h2] Solve (x-1)y''-xy'+y=0 , given x>1 and y1=e[SUP]x[/SUP] [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] I've tried solving it by multiplying everythin for (1/x-1), so my equation looks like: y''-(x/x-1)y'+y=0 (because x-1 is unequal to 0) (1) so now the equation has the form of: y''+p(x)y'+y=0, with p(x)=(-x/x-1). Then, given the fact that y[SUB]1[/SUB] is a solution, then another solution has the form Y(x)=v(x)y[SUB]1[/SUB] Then: Y'(x)=v'(x)y[SUB]1[/SUB]+v(x)y[SUB]1[/SUB]' Y''(X)=v''(x)y[SUB]1[/SUB]+v'(x)y[SUB]1[/SUB]+v'(x)y[SUB]1[/SUB]+v(x)y[SUB]1[/SUB]' Now, after replacing the y factors in (1) with Y and its derivatives, I get: v''(x)y[SUB]1[/SUB]+(2y[SUB]1[/SUB]+p(x)y[SUB]1[/SUB]=0, Which can me solved as a first order differential equation by doing v''(x)=z'(x) and v'(x)=z(x). Then, let μ(x)=e[SUP]∫p(x)dx[/SUP]=e[SUP]∫(-x/x-1)dx[/SUP]=e[SUP]-x-ln(x-1)[/SUP]=e[SUP]-x[/SUP](1/x-1) Then, z(x)=(∫μ(x)g(x)dx+c[SUB]1[/SUB])/μ(x)=c[SUB]1[/SUB]/(e[SUP]-x[/SUP]/(x-1))=c[SUB]1[/SUB](e[SUP]x[/SUP](x-1)) Now, because z(x)=v'(x)→∫z(x)dx=v(x) Then I have v(x)=c[SUB]1[/SUB]∫(e[SUP]x[/SUP](x-1)dx=c[SUB]1[/SUB]((x-2)e[SUP]x[/SUP]+c[SUB]2[/SUB]) But, whenever I do Y(x)=v(x)y[SUB]1[/SUB], and replace it in (1), it doesn't satisfy the equation- Can somebody help? [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Help Solving the following Differential equation
Back
Top