Question

Find the solution of differential equation, ^dy⁄dx = xy + x^2, if y = 1 at x = 0.

(a) \(1-\frac{x^2}{2!}+\frac{3x^4}{4!}-\frac{15x^6}{6!}+…\)

(b) \(\frac{x}{1!}+\frac{3x^3}{4!}+\frac{15x^5}{6!}+…\)

(c) \(\frac{x}{1!}-\frac{3x^3}{4!}+\frac{15x^5}{6!}-…\)

(d) \(1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+..\)

The question was asked in semester exam.

My question is based upon Taylor Mclaurin Series in chapter Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (d) \(1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+..\)

To explain: Given ^dy⁄dx = xy + x^2

hence, ^dy⁄dy (x=0) = 0

and, ^d^2y⁄dx^2= xy1 + y + 2x

hence, y2 = xy1 + y + 2x

hence, ^d^2y⁄dx^2(x=0)=1

Differentiating it n times we get,

\(y_{n+2}=xy_{n+1}+ny_n+y_n=xy_{n+1}+(n+1)y_n\)

Putting x=0 we get,

\(y_{n+2} (0)=(n+1)y_n (0)\)

Now putting the values of n as 1, 2, 3, 4, 5 we get,

\(y_3 (0)=0, \,y_4 (0)=3, \,y_5 (0)=0, \,y_6(0)=15……\) and so on

By mclaurin’s series,

\(y=1+\frac{x^2}{2!}+\frac{3x^4}{4!}+\frac{15x^6}{6!}+…\)