Recent content by jmg498
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J
Solving differential equation using the Power Series Method
That's where I'm confused. I'm not sure how to distribute the sum with that. 1 - 2/x2 = 1 - 2x-2 so does that mean the sum in the second term becomes xn-2? Not sure...- jmg498
- Post #3
- Forum: Calculus and Beyond Homework Help
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J
Solving differential equation using the Power Series Method
Homework Statement x²y'' + (x² - 2)y = 0 Homework Equations N/A The Attempt at a Solution I divided through by x² to get the equation in standard form. Then I plugged in the power series representation for y and y'' into the equation and got to this point...- jmg498
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- Differential Differential equation Method Power Power series Series
- Replies: 3
- Forum: Calculus and Beyond Homework Help
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J
Method of Undetermined Coefficients
I am looking for the form of a solution for a second order ODE with right hand side: x (sin x + 2) I'm thinking the form would be (Ax + B) sin(x) + (Cx + D) cos(x) + Ex + F. Does this seem correct? Thanks for any help or suggestions!- jmg498
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- Coefficients Method Undetermined coefficients
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Inverse Laplace Transform - Step Function?
Thanks very much! This is only my second time using TeX so I apologize for that. Getting used to it.- jmg498
- Post #6
- Forum: Calculus and Beyond Homework Help
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J
Inverse Laplace Transform - Step Function?
Ok, so I did that...here's my work...(note, I factored out the 1/10 that appears in both terms). (e^{-s}/10)\frac{17s+5}{(s+2)^{2}+4s+5}+\frac{3}{s-1} Then I completed the square on the quadratic in the denominator and rewrote it... (e^{-s}/10)\frac{17s+5}{(s+2)^{2}+1}+\frac{3}{s-1}...- jmg498
- Post #3
- Forum: Calculus and Beyond Homework Help
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Inverse Laplace Transform - Step Function?
Given... (2s^{2}+1)e^{-s} ------------------ (s - 1)(s^{2} + 4s + 5) Find the inverse Laplace Transform. I am unsure where to start and am just looking for a little direction, not an answer. From past experience, I am assuming I will have to rewrite this in another form...perhaps by...- jmg498
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- Function Inverse Inverse laplace transform Laplace Laplace transform Step function Transform
- Replies: 5
- Forum: Calculus and Beyond Homework Help
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Second Order ODE - Initial Value Problem
Ooops...added them now. Sorry! But thanks for looking!- jmg498
- Post #4
- Forum: Calculus and Beyond Homework Help
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J
Second Order ODE - Initial Value Problem
Solve the initial value problem y''+3y'+2y = 3e^{2t}+1 with initial values y(0) = 1, y'(0) = 1. I am unsure if I am going about the solution correctly. 1.) Find the characteristic equation. r^{2}+3r+2=0 \Rightarrow (r + 1)(r + 2) = 0 Therefore, y = c1•e^{-t}+c2•e^{-2t} 2.) Use method of...- jmg498
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- Initial Initial value problem Ode Second order Second order ode Value
- Replies: 4
- Forum: Calculus and Beyond Homework Help