Power series solution of&nbsp;y′−y=x2.I try to get a power series solution.y(x)=∑n=0∞anXn,y′(x)=∑n=1∞nanXn−1=∑n=0∞(n+1)an+1Xn∑n=0∞(n+1)an+1Xn−∑n=0∞anXn=X2Then&nbsp;an+1=ann+1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;∀n≠2n=2⇒3a3−a2=1Then&nbsp;a1=a0,a2=a02,a3=2+a06,a4=2+a04!,a5=2+a024⋅4Then the solution is&nbsp;y=a0+a0x1+a02x2+2+a06x3+2+a04!x4+2+a024⋅4x5⋯Is it correct? Am I supposed to find&nbsp;a0?

Question

Power series solution of&amp;nbsp;y′−y=x2.I try to get a power series solution.y(x)=∑n=0∞anXn,y′(x)=∑n=1∞nanXn−1=∑n=0∞(n+1)an+1Xn∑n=0∞(n+1)an+1Xn−∑n=0∞anXn=X2Then&amp;nbsp;an+1=ann+1&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;∀n≠2n=2⇒3a3−a2=1Then&amp;nbsp;a1=a0,a2=a02,a3=2+a06,a4=2+a04!,a5=2+a024⋅4Then the solution is&amp;nbsp;y=a0+a0x1+a02x2+2+a06x3+2+a04!x4+2+a024⋅4x5⋯Is it correct? Am I supposed to find&amp;nbsp;a0?

Jayla Matthews · Accepted Answer

Step 1Your solution amounts toy−a0(1+x+x22)=(2+a0)(ex−1−x−x22).Step 2Notice you&#039;ve indeed made a mistake. You have&amp;nbsp;an+1=ann+1, so&amp;nbsp;a5=a45=2+a05!,a6=a56=2+a06!, and so forth.Back to the topic,&amp;nbsp;y=(2+a0)ex−2(1+x+x22).It seems that any&amp;nbsp;a0&amp;nbsp;will fit&amp;nbsp;y′−y=x2. So just discard the foremost 2 and the solution is&amp;nbsp;y=a0ex−2−2x−x2.

Power series solution of y'-y=x^2. I try to get a power

Answered question

Answer & Explanation

New Questions in Calculus 2