Recent content by ChemistryNat

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    Power Series Solution to Linear ODE

    c0=1 and c1= 0? I just took the first term of the power series and it's derivative, unless I actually plug in 0 for x?
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    Power Series Solution to Linear ODE

    oh. so I need to build two series equations, how can I do this? especially since the odd terms contain a weird fraction part?
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    Power Series Solution to Linear ODE

    oh wow, the question actually asks for c1 to c4 and the radius of converge. which I assume doesn't mean I need to put the solutions into series form.. my apologies. The radius of convergence is |x-x0|<R but how do you find R?
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    Power Series Solution to Linear ODE

    So all of my odd terms will be expressed in terms of c1 and all the even terms c0 which means I need two solutions? I'm not very familiar with power series in general, I struggled with them in first year calc too
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    Power Series Solution to Linear ODE

    $$ c_5=\frac{8c_1 - 14}{20} $$ which I think I'm supposed to see some sort of condensed version involving n! of some sort? and $$ c_6=\frac{13c_4 + 2c_2}{30} $$ but I have two equations with C2 and one with both c2 and c4?
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    Power Series Solution to Linear ODE

    oh wow! so many mistakes! Thank you So for the recurrence relation, I want it in terms of cn+2 $$ c_{n+2} = \frac{[n(n-1)+1]c_n + (n-2)c_{n-2}}{(n+2)(n+1)} = 0 $$ and then I solve for n=3 giving me c5= $$ \frac{7c_3 c_1}{20} $$ ?
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    Power Series Solution to Linear ODE

    Sorry, it would seem I made an error on the first line, but after that I think this term has been corrected
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    Laplace Transform Solution to Second Order ODE IVP

    definitely an algebra error present, Y(s)(s2+6)+2=1/s2-e-s/s2-e-s/s then bring the 2 over Y(s)(s2+6)=1/s2-e-s/s2-e-s/s-2 divided out the (s2+6) term Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)
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    Power Series Solution to Linear ODE

    Homework Statement Let y(x)=\sumckxk (k=0 to ∞) be a power series solution of (x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0 Note that x=0 is an ordinary point. Homework Equations y(x)=\sumckxk (k=0 to ∞) y'(x)=\sum(kckxk-1) (k=1 to ∞) y''(x)=\sum(k(k-1))ckxk-2 (k=2 to ∞) The Attempt at a Solution...
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    Second Order Linear ODE - Power Series Solution to IVP

    Oh! so I can just leave it be and put it with the other x1 coefficient terms?
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    Second Order Linear ODE - Power Series Solution to IVP

    So I can leave it in as a constant coefficient term, say 2C1x? and then use that in combination with the other constants I've pulled out?
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    Second Order Linear ODE - Power Series Solution to IVP

    Homework Statement Let y(x)=\sumckxk (k=0 to ∞) be a power series solution of (x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0 Note that x=0 is an ordinary point. Homework Equations y(x)=\sumckxk (k=0 to ∞) y'(x)=\sum(kckxk-1) (k=1 to ∞) y''(x)=\sum(k(k-1))ckxk-2 (k=2 to ∞) The Attempt at a Solution...
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    Proving Subspace of ℝ^{n} from Linear Algebra Homogeneous System

    That x+y is also a solution? It should be if it's closed under addition
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    Linear Algebra Basis for Hyperplane - Example 8 Explanation

    Am I making a solution vector that includes the variables x1, x2, x3 and x4?
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    Proving Subspace of ℝ^{n} from Linear Algebra Homogeneous System

    Set is a subspace if it is closed under addition and scalar multiplication (and therefore include the zero vector)