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 March 2nd, 2011, 03:07 AM #1 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Differential equations Find the general solution of equation d^2y/dx^2= 12x^2 I got: dy/dx=12x^3/3 + A ...integrate again =y=x^4+Ax+B ---> General solution Then, find solution which satisfy 1) y(0)=2, y(1)=8 given y(0)=2 2=0+A(0)+B ---> B=2 given y(1)=8 8=1+A(1)+2 ----> A=8-2-1 A=5 So. y=x^4+5x+2 ---> Particular solution does this look correct to you? 2) y(0)=1, y'(0)=-2 .... does the mark above the second y just mean the derivative of y? and what would I do with this? thanks
 March 2nd, 2011, 03:18 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Differential equations 1) Yes 2) Yes; y' = 4x³+A, so y'(0) = -2 implies A = -2.
 March 2nd, 2011, 04:00 AM #3 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations cheers buddy
 March 3rd, 2011, 10:48 AM #4 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Hey, I've got 2 equations to solve but I don't know how to go about solving them as we haven't covered the topic in depth yet. Any help would be appreciated. 1) dy/dx=e^-x/y 2) dy/dx=6sinx/y Jake
 March 3rd, 2011, 11:13 AM #5 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Does u = x/y substitution work?
 March 3rd, 2011, 06:11 PM #6 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 The problems are a bit ambiguous as typed. If they are: 1) $\frac{dy}{dx}=\frac{e^{-x}}{y}$ 2) $\frac{dy}{dx}=\frac{6\sin(x)}{y}$ Then these are what's known as first order separable differential equation. Separable means the variables may be separated to either side of the equation, and may be solved directly by integration. 1) may be separated as follows: $y\,dy=e^{-x}\,dx$ $\int y\,dy=\int e^{-x}\,dx$ $\frac{1}{2}y^2=-e^{-x}+C$ 2) may be separated as follows: $y\,dy=6\sin(x)\,dx$ $\int y\,dy=6\int\sin(x)\,dx$ $\frac{1}{2}y^2=-6\cos(x)+C$ You may check these solutions through implicit differentiation. If they are: 1) $\frac{dy}{dx}=e^{-\frac{x}{y}}$ 2) $\frac{dy}{dx}=6\sin(\frac{x}{y}\)$ then the substitution suggested by [color=#008000]johnny[/color] will make the equations separable, however, this doesn't guarantee they will be integrable, i.e., the resulting integrands may not have anti-derivatives expressible in elementary terms. Suppose we have the first order ODE $\frac{dy}{dx}=g$$\frac{x}{y}$$$ Let $u=\frac{x}{y}\:\therefore\:y=\frac{x}{u}$ Noting that both y and u are functions of x, we have from the quotient rule: $\frac{dy}{dx}=\frac{u-x\frac{du}{dx}}{u^2}$ Substituting for dy/dx into the ODE, we have: $\frac{u-x\frac{du}{dx}}{u^2}=g(u)$ This is a separable equation, and may be written: $\frac{1}{u$$1-u\cdot g(u)$$}\,du=\frac{1}{x}\,dx$ We obtain the implicit solution from: $\int\frac{1}{u$$1-u\cdot g(u)$$}\,du=\int\frac{1}{x}\,dx$ I suspect, from the given functions of u, resulting in non-linear equations, that my first interpretation is correct.
March 3rd, 2011, 08:35 PM   #7
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Quote:
 Originally Posted by MarkFL The problems are a bit ambiguous as typed. If they are: 1) $\frac{dy}{dx}=\frac{e^{-x}}{y}$ 2) $\frac{dy}{dx}=\frac{6\sin(x)}{y}$ Then these are what's known as first order separable differential equation. Separable means the variables may be separated to either side of the equation, and may be solved directly by integration. 1) may be separated as follows: $y\,dy=e^{-x}\,dx$ $\int y\,dy=\int e^{-x}\,dx$ $\frac{1}{2}y^2=-e^{-x}+C$ 2) may be separated as follows: $y\,dy=6\sin(x)\,dx$ $\int y\,dy=6\int\sin(x)\,dx$ $\frac{1}{2}y^2=-6\cos(x)+C$ You may check these solutions through implicit differentiation.
Good job!

 March 4th, 2011, 01:00 AM #8 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Differential equations Im pretty sure your solution is correct as it is similar to an example i was given, it seems easier to work using the seperation method. thanks So, with my next question. ''Solve the equation subject to the condition y(0)=1 dy/dx= 3x^2 e^-y '' I would solve the equation and then plug in the boundary condition value?
 March 4th, 2011, 10:39 AM #9 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, or you can simply use the initial and final conditions as the limits of integration to eliminate the need to solve for the constant of integration. Suppose you have the IVP: $\frac{dy}{dx}=f(x)$ , $y\left(x_0\right)=y_0$ Now, using indefinite integrals, we have: $\int\,dy=\int f(x)\,dx$ $y(x)=F(x)+C$ Using the initial condition, we get $C=y_0-F(x_0)$ thus: $y(x)=F(x)+y_0-F(x_0)$ which we may rewrite as: $y(x)-y_0=F(x)-F(x_0)$ which we may rewrite, using the anti-derivative form of the fundamental theorem of calculus as $\int_{y_0}\,^{y(x)}\,dy=\int_{x_0}\,^{x}f(x)\,dx$ Using the given problem, I'll demonstrate both methods to show they give the same result. (i) $\frac{dy}{dx}=3x^2e^{-y}$ Separate variables, and integrate: $\int e^y\,dy=\int 3x^2\,dy$ $e^y=x^3+C$ Convert from exponential to logarithmic form: $y=\ln\|x^3+C\|$ Use initial condition to find C: $1=\ln\|0^3+C\|\:\therefore\:C=e$ so our solution is: $y=\ln\|x^3+e\|$ ii) Here, when we separate variables and integrate, we'll use definite integrals, with the boundaries as the limits of integration: $\int_1\,^y e^y\,dy=\int_0\,^x 3x^2\,dy$ $e^y-e=x^3$ $e^y=x^3+e$ $y=\ln\|x^3+e\|$ So, the method you use is simply a matter of preference.
 March 5th, 2011, 02:01 AM #10 Newbie   Joined: Dec 2010 Posts: 11 Thanks: 0 Re: Differential equations Awesome Finding the general solution of the equation dx/dt=t(x-2) hence find the particular solution when x(0)=5 When I try integrate the term wrt "t", I must be doing it wrong as I have only tried previous to this finding the general solution to a second order ODE, which this isn't.

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