Thread:

1. Solving differential equations using a change of variable

My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation?

2. Erghhhhhhh....:p

Originally Posted by Erghhh
My book is telling me

d/du((e^u)dy/dx)

=(e^u)dy/dx + (e^u)(d2y/dx2).(dx/du)

I understand where the first part has come from, but not sure how the second part was derived. Would anyone kindly provide me with an explanation?
This will make it clear

$\frac{d}{du}\{e^u\frac{dy}{dx}\}$

$= \frac{dy}{dx}\frac{d}{du}(e^u) +e^u \frac{d}{du}(\frac{dy}{dx})$

Differnetiation of dy/dx wrt u will be given by (using chain rule)

$= \frac{y'}{dx} \cdot \frac{du}{dx}$
------------------------------------------
So differnetiation will be

$=\frac{dy}{dx}e^u + e^u \frac{d^2(y)}{dx^2} \cdot \frac{dx}{du}$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9249181151390076, "perplexity_flag": "middle"}