Practice Final Solutions

# Practice Final Solutions - Math 127 S2008 Solutions for the...

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Unformatted text preview: Math 127 - S2008 Solutions for the Practice Final 1. Show that the function y = C- cos( x ) x 2 is a solution of the differential equation x 2 y + 2 xy = sin( x ) . For what value of C does the solution satisfy the initial condition y (2 π ) = 0? Solution: y = x sin( x )- 2( C- cos( x )) x 3 and thus x 2 y + 2 xy = x 3 sin( x )- 2 x 2 ( C- cos( x )) x 3 + 2 x ( C- cos( x )) x 2 = sin( x )- 2 x ( C- cos( x )) + 2 x ( C- cos( x )) = sin( x ) . 2. Find the general solution to the differential equation y = xy 4 e 3 x 2 . Solution: We have Z dy y 4 = Z xe 3 x 2 dx. First Z dy y 4 =- 1 y 3 + C 1 and then Z xe 3 x 2 dx = e 3 x 2 6 + C 2 . Thus- 1 y 3 = e 3 x 2 6 + C. (This is enough for the final answer, but you can also solve for y if you like, and get y =- e 3 x 2 6 + C !- 1 / 3 . 3. Solve the initial value problem dX dt = sin( t ) s X 3 cos( t ) , X (0) = 4 . Solution: We have Z dX X 3 / 2 = Z sin ( t ) dt p cos( t ) , 1 which has a solution- 2 √ X =- 2 p cos( t ) + C....
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Practice Final Solutions - Math 127 S2008 Solutions for the...

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