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 March 14th, 2011, 12:19 PM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Differential equations hey, wondering if this equation I should use substitution in the process or a different method. ie u=x/t find the general solution of the equation : (xt)dx/dt = x^2 + t^2 thanks
 March 14th, 2011, 12:40 PM #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: Differential equations Yes, that is an appropriate substitution for this non-linear equation, leading to an implicit solution for x(t).
 March 15th, 2011, 10:41 AM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Can you check a these please 1. Find the general solutions of the first order equations. a) y dy/dx=x+2 ?y dy =?(x+2) dx y^2/2 = (1/2 x^2 +2x) +C y^2 = x^2 +4x + C y=sqrt (x^2 +4x + C) general solution b) dy/dx (y^2 -1) / x (1/y^2) -1 dy/dx = 1/x ?(1/y^2) -1 dy = ?1/x (1/y^2 -1) dy =?ln(x) +C y=tan(ln(x)+C) general solution c)x^2 dy/dx +2xy=2x d/dx(x^2 y)=2x exact form ?d/dx (x^2 y) dx = ?2x dx x^2 y = x^2 +C y=x^2 / x^2 + C/x^2 y=1+Cx^-2 general solution d) cosx dy/dx -y sinx = 1 d/dx (cosx y) =1 exact form ?d/dx (cosx y) dx =?1 dx cosx y = x y= x/cosx + C general solution
 March 15th, 2011, 11:17 AM #4 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: Differential equations a) You are correct up to the point: $y^2=x^2+4x+C$ Next I would write it as: $y^2-$$x^2+4x$$=C$ Complete the square on x: $y^2-$$x^2+4x+4$$=C$ $y^2-$$x+2$$^2=C$ $\frac{y^2}{C}-\frac{$$x+2$$^2}{C}=1$ This is the family of "north-south" opening square hyperbolas centered at (-2,0) When you take the square root of both sides of an equation you need to put a ± on one side, typically the side opposite the variable you are solving for. As you wrote the solution, you are only getting the top-half of the hyperbolas in the solution space. b) $\frac{dy}{dx}=\frac{y^2-1}{x}$ Separate variables: $\frac{1}{y^2-1}\,dy=\frac{1}{x}\,dx$ Use partial fraction decomposition on the left side: $\frac{1}{2}\int \frac{1}{y-1}-\frac{1}{y+1}\,dy=\int \frac{1}{x}\,dx$ $\ln\|\frac{y-1}{y+1}\|=\ln|Cx^2|$ $\frac{y-1}{y+1}=Cx^2$ $y-1=yCx^2+Cx^2$ $y$$1-Cx^2$$=1+Cx^2$ $y=\frac{1+Cx^2}{1-Cx^2}$ c) Correct. d) You need to put your constant of integration in place before you divide through by cos(x): $y\cos(x)=x+C$ $y=$$x+C$$\sec(x)$
 March 15th, 2011, 12:18 PM #5 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Thanks very much , although I was sure I had done well on those, never mind, your steps are good to follow.
 March 15th, 2011, 12:40 PM #6 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Ok, so i attempted my next question, hopefully i did better on this one. Find the particular solution of the first order differential equation x^2 dy/dx -y = 1 subject to y(2)=2 divide by x^2 to get equation in standard form dy/dx - 1/x y = 1/x^2 ?(x)=e^-?1/x dx =e^-?x^-2 = e^x^-1 multiply through by integrating factor e^x^-1 dy/dx - x^-2 y e^x^-1 = 1/x^2 e^x^-1 d/dx (e^x^1 y) = 1/x^2 e^x^-1 ?d/dx (e^x^-1 y) dx = ?1/x^2 e^x^-1 dx e^x^-1 y = 1/x^2 e^1/x e^1/x y = x^-2 e^1/x y= x^-2 e^1/x +C / e^1/x y=x^-2 = 1/x^2 + c/e^1/x general solution y(2)=2 2=1/2^2 + c/e^1/2 2=0.25+c/1.6487 (2-0.25) x 1.6487 =c 1.75 x 1.6487 = c 2.8358 =c y=1/x^2 + 2.8358 / e^1/x particular solution
 March 15th, 2011, 03:17 PM #7 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: Differential equations You have: $x^2\frac{dy}{dx}-y=1$ Putting the equation in standard form, you should get: $\frac{dy}{dx}+$$-\frac{1}{x^2}$$y=\frac{1}{x^2}$ Calculate integrating factor: $\mu(x)=e^{-\int x^{-2}\,dx}=e^{\frac{1}{x}}$ You have the correct integrating factor, so I assume your equation in standard form just contains a typo, where you have P(x) = 1/x. $e^{\frac{1}{x}}\frac{dy}{dx}+e^{\frac{1}{x}}$$-\frac{1}{x^2}$$y=\frac{1}{x^2}e^{\frac{1}{x}}$ $\frac{d}{dx}$$e^{\frac{1}{x}}y$$=\frac{1}{x^2}e^{\ frac{1}{x}}$ When you integrate the right side, you need to see that you initially have $e^{u(x)}$$-\frac{du}{dx}$$$ so you need to multiply the integral by a negative so that you get: $-\int e^u\,du=-e^u+C$ In our case, $u=\frac{1}{x}:\therefore\:du=-\frac{1}{x^2}\,du$ $\int\frac{d}{dx}$$e^{\frac{1}{x}}y$$\,dx=-\int\ \frac{d}{dx}$$e^{\frac{1}{x}}$$\,dx$ $\int\,d(e^{\frac{1}{x}}y\)=-\int\,d$$e^{\frac{1}{x}}$$$ $e^{\frac{1}{x}}y=-e^{\frac{1}{x}}+C$ $y=Ce^{-\frac{1}{x}}-1$ Determine C from initial conditions: $2=Ce^{-\frac{1}{2}}-1$ $C=3e^{\frac{1}{2}}$ Thus the solution satisfying the given conditions is: $y=3e^{\frac{1}{2}}e^{-\frac{1}{x}}-1=3e^{\frac{x-2}{2x}}-1$
 March 15th, 2011, 05:18 PM #8 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: Differential equations By the way, here is the working of the problem you gave in the first post: $xt\frac{dx}{dt}=x^2+t^2$ Divide through by xt: $\frac{dx}{dt}=\frac{x}{t}+\frac{t}{x}$ Use the substitution: $u=\frac{x}{t}\:\therefore\=ut\:\therefore\:\frac {dx}{dt}=u+t\frac{du}{dt}" /> The equation is transformed to: $u+t\frac{du}{dt}=u+\frac{1}{u}$ $t\frac{du}{dt}=\frac{1}{u}$ $u\,du=\frac{1}{t}\,dt$ $\int u\,du=\int \frac{1}{t}\,dt$ $\frac{1}{2}u^2=\ln|Ct|$ $u^2=\ln$$\(Ct$$^2\)$ $x^2=t^2\ln$$\(Ct$$^2\)$
 March 16th, 2011, 03:26 AM #9 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Great what about this one.. x dy/dx -2y = x^4 e^x divide by x^3 x/x^3 dy/dx -2y/x^3 = x^4 e^x/ x^3 x^-2 dy/dx -2x^-3 y = xe^x d/dx (x^-2 y) =xe^x exact form integrate wrt x d/dx(x^-2 y) dx = xe^x dx x^2 y = xe^x + e^x y= xe^x + e^x / x^2 y=e^x/x + e^x/x^2 general solution
 March 16th, 2011, 05:48 AM #10 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: Differential equations When you integrate: $\int\,d$$\frac{y}{x^2}$$=\int xe^x\,dx$ You get: $\frac{y}{x^2}=xe^x-e^x+C=e^x(x-1)+C$ $y=x^2$$e^x(x-1)+C$$$

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