Solve the following initial-value problems. For and show by substitution that yo

Question

Solve the following initial-value problems. For and show by substitution that your solutions are correct (both the equation and the initial condition - to get full marks you MUST do this). Of course you should ALWAYS check your solutions. In (c) you will not be able to solve the resulting integral. Use Wolfram Alpha [http://www.wolframalpha. com] (see below for some help) or another package (e.g. Mat lab symbolic toolbox) to write the solution and find the value of the constant from the boundary condition - to do this you will need to find out something about the integral known as the Error Function - Erf(x).

Explanation / Answer

b) y' - y/2 = e^(x/2) sinx, y(0) = 1 y' + P(x) y = Q(x) linear diff. eqn integration factor i(x) = e^(int P(x) dx ) here int means integral i(x) = e(int (-1/2) dx) = e^(-x/2) i(x) y = int (i(x) Q(x)) dx + C e^(-x/2) y = int ( e^(-x/2) e^(x/2) sinx ) dx + C e^(-x/2) y = int ( sinx ) dx + C e^(-x/2) y = - cosx + C y= e^(x/2) (- cosx + C) y(0) = 1==> 1= 1( cos0 +C ) ==> 1= 1( 1 + C) ==> C=0 y= - e^(x/2) cosx Check!!!! y= - e^(x/2) cosx y' - y/2 = e^(x/2) sinx y' = -(1/2) e^(x/2) cosx - ( - e^(x/2) sinx ) y' = -(1/2) e^(x/2) cosx + e^(x/2) sinx y= - e^(x/2) cosx ==> -(1/2) y = (1/2) e^(x/2) cosx y' - (1/2) y = -(1/2) e^(x/2) cosx + e^(x/2) sinx + (1/2) e^(x/2) cosx y' - (1/2) y = e^(x/2) sinx THat is true a) y' = -2y^2 lnx, y(1) = 1/2 dy/dx = -2y^2 lnx dy / y^2 = -2 lnx dx integrate both sides -1/y = -2 ( x ln x - x ) + C y(1) = 1/2 ==> - 1/1 = -2 ( 1 ln 1 - 1 ) + C -1= -2(0 -1) + C -1 = 2 +C C= -3 -1/y = -2 ( x ln x - x ) + C -1/y = -2 ( x ln x - x ) -3 Check - y^(-1) = -2 ( x ln x - x ) -3 y^(-2) y' = -2( lnx+x/x -1) - 0 y' /y^2 = -2( lnx) y' = -2 y^2 lnx THat is true

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