document.write( "Question 358398: Solve the following initial value problems:
\n" ); document.write( "A)x\"+x=0 :x(0)=0,x'(0)=1\r
\n" ); document.write( "\n" ); document.write( "b)x\"+4x=0; x(0)=1, x'(0)=0 \n" ); document.write( "

Algebra.Com's Answer #255796 by robertb(5830) $\"\"$ $\"About$

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A) We can let $\"x%28t%29+=+AsinBt\"$ . This gives $\"dx%2Fdt+=+AB%2A%28cosBt%29\"$ , and $\"d%5E2x%2Fdt%5E2+=+-AB%5E2%2A%28sinBt%29\"$ . Then
\n" ); document.write( " $\"d%5E2x%2Fdt%5E2+%2B+x+=+AsinBt-AB%5E2%2A%28sinBt%29+=+A%2A%281-B%5E2%29%2AsinBt\"$ . This implies that $\"A%281-B%5E2%29+=+0\"$ , from the given DE. Since we don't want A = 0, then B = 1. (We can drop B = -1 without loss of generality.)
\n" ); document.write( "So $\"x%28t%29+=+Asint\"$ . From the initial value x'(0) = 1, we get A = 1. Therefore $\"x%28t%29+=+sint\"$ .
\n" ); document.write( "B) The solution of this IVP follows along the same line as above, but this time we let $\"x%28t%29+=+AcosBt\"$ . By using the differential equation we get B=2, and by using the initial condition, we get A = 1. Therefore $\"x%28t%29+=+cos2t\"$ . \n" ); document.write( "

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