Question

Find the first five nonzero terms in the solution of the given initial value problem. y" + xy + 2y = 0, y (0) = 4, y' (0) = 7 Enter an exact answer y =

Answer 1

Solution:-

Given that

$y''+xy'+2y=0$

,

= (0)

L = (0),

$y''+xy'+2y=0$

$y=\sum^\infty_{n=0}a_nx^n$

$y'=\sum^\infty_{n=1}na_nx^{n-1}$

$y''=\sum^\infty_{n=2}n(n-1)a_nx^{n-2}$

put these values in given diff equation

$\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+x\sum^\infty_{n=1}na_nx^{n-1}+2\sum^\infty_{n=0}a_nx^n=0$

$\sum^\infty_{n=2}n(n-1)a_nx^{n-2}+\sum^\infty_{n=1}na_nx^n+2\sum^\infty_{n=0}a_nx^n=0$

$\sum^\infty_{n=0}(n+2)(n+1)a_{n+2}x^n+\sum^\infty_{n=0}na_nx^n+2\sum^\infty_{n=0}a_nx^n=0$

$\sum^\infty_{n=0}[(n+2)(n+1)a_{n+2}+na_n+2a_n]x^n=0$

$\sum^\infty_{n=0}[(n+2)(n+1)a_{n+2}+(n+2)a_n]x^n=0$

$(n+2)(n+1)a_{n+2}+(n+2)a_n=0$

$(n+2)(n+1)a_{n+2}=-(n+2)a_n$

$\therefore a_{n+2}=-\frac{1}{(n+1)}a_n$

$y=\sum^\infty_{n=0}a_nx^n$

$y(x)=a_0+a_1x+a_2x^2+a_3x^3+...$

$y'(x)=a_1+2a_2x+...$

Given

= (0)

$a_0=4$

L = (0),

$a_1=7$

$a_{n+2}=-\frac{1}{(n+1)}a_n$

n = 0

$a_{2}=-\frac{1}{1}a_0=-4$

$a_{2}=-4$

n = 1

$a_{3}=-\frac{1}{2}a_1$

$a_{3}=-\frac{1}{2}*7$

$a_{3}=-\frac{7}{2}$

n = 2

$a_{4}=-\frac{1}{3}a_2$

$a_{4}=-\frac{1}{3}(-4)$

$a_{4}=\frac{4}{3}$

$y(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$

$y(x)=4+7x-4x^2-\frac{7}{2}x^3+\frac{4}{3}x^4+...$

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