# A differential equation question

If $\frac{dy}{dx} = \sqrt{y^2+1}$, then $\frac{d^2y}{dx^2}$=?
The correct answer is $y$.

-

It's just a few applications of the chain rule and use of the ODE as $\frac d{dx} y(x)$ appears. \begin{align*} \frac d{dx} \sqrt{y^2(x)+1} & = \frac1{2\sqrt{y^2(x)+1}} \cdot \frac d{dx} (y^2(x) + 1) \\ & = \frac1{2\sqrt{y^2(x)+1}} 2y(x) \cdot y'(x) \\ & \stackrel{\text{ODE}}= \frac1{\sqrt{y^2(x)+1}} \cdot y(x) \cdot \sqrt{y^2(x)+1} \\ & = y(x) \end{align*}
Hint: $\frac{d^{2}y}{dx^2} = \left( \frac{dy}{dx}\right)'$
Hint: Use chain rule while differentiating $\frac{dy}{dx}$ with respect to $x$.
{"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000100135803223, "perplexity": 1635.3724945480174}, "config": {"markdown_headings": true, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00088-ip-10-185-27-174.ec2.internal.warc.gz"}