Questions on basic differential equations

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SUMMARY

This discussion focuses on solving basic differential equations as part of a homework assignment. The first problem involves deriving the initial value problem for an object released from a height, leading to the equation Vo=(2h-gTo^2)/(2To). The second problem requires finding a solution to the differential equation (1-x^2)y'' - xy' + 4y = 0 using a polynomial form Y(x) = a0 + a1x + a2x^2. The third problem involves determining the differential equation for the slope of the tangent line for the curve defined by x^2 + y^2 = 2cx. Participants provide step-by-step guidance on how to approach these problems.

PREREQUISITES
  • Understanding of basic differential equations
  • Familiarity with initial value problems
  • Knowledge of polynomial solutions to differential equations
  • Ability to differentiate implicit functions
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  • Study the method of solving initial value problems in differential equations
  • Learn about polynomial solutions and the method of undetermined coefficients
  • Explore implicit differentiation techniques for curves
  • Review the concepts of normalization conditions in differential equations
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Students studying differential equations, educators teaching introductory calculus, and anyone seeking to improve their problem-solving skills in mathematical physics.

danield
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Well this is my first diff eqs homework and I am totally lost, i have no idea what to do here are the questions that i have...

Homework Statement


1)An object released from a height h meters above the gorund with a veritcal velocity of Vo m/s htis the ground after To seconds. Neglecting fricitonal forces set up and solve the inital value problem governing the motion and use your solution to show that...
Vo=(2h-gto^2)/(2To)

2) Determine a solution to the differential equation
(1-x^2)y'' - xy' +4y = 0
of the form Y(x) = a0 + a1x + a2x^2 satisfying the normalization condition y(1) = 1

3) Determine the differential equaiton giving the slope of the tangent line at the point (x,y) for the given curve
x^2 + y^2 = 2cx


The Attempt at a Solution


1)on that one i got till the point where i have
1/2gt^2 + C1t+ c2 = y(t)


2) On this one i have no idea how to start maybe help on where to start will be enough...
3) on this one i also have no idea where to start...

im still working on other ones ill see if I am able to work them out


any help is appreciated
thanks
 
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danield said:
Well this is my first diff eqs homework and I am totally lost, i have no idea what to do here are the questions that i have...

Homework Statement


1)An object released from a height h meters above the gorund with a veritcal velocity of Vo m/s htis the ground after To seconds. Neglecting fricitonal forces set up and solve the inital value problem governing the motion and use your solution to show that...
Vo=(2h-gto^2)/(2To)

2) Determine a solution to the differential equation
(1-x^2)y'' - xy' +4y = 0
of the form Y(x) = a0 + a1x + a2x^2 satisfying the normalization condition y(1) = 1

3) Determine the differential equaiton giving the slope of the tangent line at the point (x,y) for the given curve
x^2 + y^2 = 2cx


The Attempt at a Solution


1)on that one i got till the point where i have
1/2gt^2 + C1t+ c2 = y(t)
Okay, you know that y(0)= (1/2)g(0^2)+ C1(0)+ C2= h (An object released from a height h meters above the ground) and that y'(0)= g(0)+ C1= V0 (with a vertical velocity of Vo) so you can determine C1 and C2 from those.

2) On this one i have no idea how to start maybe help on where to start will be enough...
Do it! You are asked for a solution of the form [itex]y= a_0+ a_1x+ a_2x^2[/itex] so [itex]y'= a_1+ 2a_2x[/itex] and [itex]y"= 2a_2[/itex]. Put those into the differential equation and solve for [itex]a_0[/itex], [itex]a_1[/itex], and [itex]a_2[/itex] by, for example, choosing three values for x to get three equations.

3) on this one i also have no idea where to start...
You have a "family" of curves given by [itex]x^2+ y^2= 2cx[/itex] and you want an equation involving y' and/or y" but not c. Now, you could just differentiate with respect to x twice and that would eliminate c: [itex]2x+ 2y y'= 2c[/itex] and then [itex]2+ 2(y')^2+ 2y y"= 0[/itex]. But since there is only one "c" you should be able to do this with just one differentiation. Differentiating once gives [itex]2x+ 2y y'= 2c[/itex] so [itex]c= x+ y y'[/itex]. Replace c in [itex]x^2+ y^2= 2cx[/itex] with that!

im still working on other ones ill see if I am able to work them out


any help is appreciated
thanks
 

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