My Math Forum Differentiation trouble

 Calculus Calculus Math Forum

 March 20th, 2012, 11:56 PM #1 Member   Joined: Feb 2012 Posts: 44 Thanks: 0 Differentiation trouble I am having trouble with these: -y = 1/(3 + x^2)<--denominator is under square root. -dy/dx: -x/(y(x)^3) -y(x)^2 + x^4y'(x)^2 = 1 <--the second y is prime -y''(x) + 2y'(x) + 5y(x) = 0 <-- The first y is 2 prime and the second y is prime I have a final on monday and am trying to understand these the best I can P.S. Does anybody know of a site that can teach me how to express these equations on the computer to make them look authentic to the original?
 March 21st, 2012, 05:32 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Differentiation trouble 1.) $y=\frac{1}{\sqrt{3+x^2}}=$$x^2+3$$^{\small{-\frac{1}{2}}}$ To differentiate, use the power and chain rules: $\frac{dy}{dx}=-\frac{1}{2}$$x^2+3$$^{\small{-\frac{3}{2}}}\cdot$$2x$$=-\frac{x}{$$x^2+3$$^{\small{\frac{3}{2}}}}$ 2.) $\frac{dy}{dx}=-\frac{x}{y^3(x)}$ Separate variables and integrate: $\int y^3(x)\,dy=-\int x\,dx$ $\frac{y^4(x)}{4}=-\frac{x^2}{2}+C$ The solution is given implicitly by: $y^4(x)+2x^2=C$ 3.) $y^2(x)+x^4y'^2(x)=1$ Separate variables and integrate: $\int\frac{1}{\sqrt{1-y^2(x)}}\,dy=\pm\frac{1}{x^2}$ $\sin^{\small{-1}}$$y(x)$$=\mp\frac{1}{x}+C$ $y(x)=\sin$$\mp\frac{1}{x}+C$$$ 4.) $y''(x)+2y'(x)+5y(x)=0$ This is a second order homogeneous equation whose associated auxiliary equation has the roots: $r=-1\pm2i$ Hence, the general solution is given by: $y(x)=e^{-x}$$c_1\cos(2x)+c_2\sin(2x)$$$ Do a search here and on the web for LaTeX to see how to write math expressions.

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