حل معادله خطی مرتبه دوم با اموزش کامل

[hamidrezavictor]

با سلام حل معادله خطی مرتبه دوم ...........................( y"+2/x y'+1/x^4 y=1................y1=sin(1/x..............

[hts1369]

با سلام حل معادله خطی مرتبه دوم ...........................( y"+2/x y'+1/x^4 y=1................y1=sin(1/x..............

ای کاش کمی توضیح داده بودی
منظورت از y1 یکی از جوابهاست و جواب دوم رو میخاهی؟؟

[hamidrezavictor]

ای کاش کمی توضیح داده بودی
منظورت از y1 یکی از جوابهاست و جواب دوم رو میخاهی؟؟[ برای مشاهده لینک ، لطفا با نام کاربری خود وارد شوید یا ثبت نام کنید ] XNSR0IArs4c6QAAAARnQU1BAACxjwv8YQUAAAAJcEhZcwAADsM AAA7DAcdvqGQAADuCSURBVHhe7Z0v1B63sYfTU9Ce03N6DAsDC wMDDQ0NAz9oGGhmGGhoaGhoGBhoGGgYaGgY2Dv36kZRd/fd1f+Z0T4GPW4sjUbPaPX+dqSVvvkPfyAAAQhAAAIQgAAESgh8 U1KYshCAAAQgAAEIQAAC/0E/MQggAAEIQAACEIBAGQH0UxkvSkMAAhCAAAQgAAH0E2MAAhCAAA QgAAEIlBFAP5XxojQEIAABCEAAAhBAPzEGIAABCEAAAhCAQBkB 9FMZL0pDAAIQgAAEIAAB9BNjAAIQgAAEIAABCJQRQD+V8aI0BC AAAQhAAAIQQD8xBiAAAQhAAAIQgEAZAfRTGS9KQwACEIAABCAA AfQTYwACEIAABCAAAQiUEUA/lfGiNAQgAAEIQAACEEA/MQYgAAEIQAACEIBAGQH0UxkvSkMAAhCAAAQgAAH0E2MAAhCAAA QgAAEIlBFAP5XxojQEIAABCEAAAhBAPzEGIAABCEAAAhCAQBkB 9FMZL0pDAAIQgAAEIAAB9BNjAAIQgAAE3BP4+eefnz9//v79e/c9oQNOCKCfnAQKNyEAAQhA4DGBf/3rX998883f//73tAiiiiEzjgD6aRxbLEMAAhCAwCQCIp7Cn7S9Q1E1ySGaWZ0A 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این معادله خطی مرتبه دوم غیر همگن با جواب عمومی y=yg + yp که yg همان جواب عمومی معادله همگن و yp عبارت است از y2=V1y1+V2y2
V1 و V2 توابعی هستند که مشتقاتشون طبق فرمول زیر و انتگرال گیری بدست میاد

[hamidrezavictor]

فرمول اینجا اپلود کردم [ برای مشاهده لینک ، لطفا با نام کاربری خود وارد شوید یا ثبت نام کنید ]

[hamidrezavictor]

y1 هم برای گذاشتن تو فرمول معمولا خودشون میدن

[hts1369]

بله بلد هستم این نوع معادله رو
بفرمایید ([ برای مشاهده لینک ، لطفا با نام کاربری خود وارد شوید یا ثبت نام کنید ](\frac{1}{x}\right)\int&space;\f rac{1}{\sin&space;^2\left(\frac{1}{x}\right)}e^{-\int&space;\frac{2}{x}dx}dx&space;\\&space;y_2=\sin&space;\left(\frac{1}{x }\right)\int&space;\frac{1}{x^2\sin&space;^2\left(\frac{1}{x}\ right)}dx&space;\\&space;\frac{1}{x}=u\Rightarrow&space;\frac{-1}{x^2}dx=du&space;\\&space;y_2=\sin&space;(u)\int&space;\frac{-1}{\sin&space;^2(u)}du=\sin&space;(u)\left(\int&space;-\left(1+\cot&space;^2u\right)du\right)=\sin&space;(u)(\cot&space;u)= \cos&space;u=\cos&space;\left(\frac{1}{x}\right))

[hamidrezavictor]

بله بلد هستم این نوع معادله رو
بفرمایید ([ برای مشاهده لینک ، لطفا با نام کاربری خود وارد شوید یا ثبت نام کنید ](\frac{1}{x}\right)\int&space;\f rac{1}{\sin&space;^2\left(\frac{1}{x}\right)}e^{-\int&space;\frac{2}{x}dx}dx&space;\\&space;y_2=\sin&space;\left(\frac{1}{x }\right)\int&space;\frac{1}{x^2\sin&space;^2\left(\frac{1}{x}\ right)}dx&space;\\&space;\frac{1}{x}=u\Rightarrow&space;\frac{-1}{x^2}dx=du&space;\\&space;y_2=\sin&space;(u)\int&space;\frac{-1}{\sin&space;^2(u)}du=\sin&space;(u)\left(\int&space;-\left(1+\cot&space;^2u\right)du\right)=\sin&space;(u)(\cot&space;u)= \cos&space;u=\cos&space;\left(\frac{1}{x}\right))

ممنونم لطف کردین ... :n01: میشه چندتا جزوه اموزشی معادلات هم معرفی کنین برای یادگیری بهتر ممنون میشم ازتون

[hts1369]

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