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# 0005 - This section introduces general solutions and...

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Unformatted text preview: This section introduces general solutions and particular solutions in the very simplest situation — a differential equation of the form y'= f (x)— where only direct integration and evaluation of the constant of integration are involved. Students should review carefully the elementary concepts of velocity and acceleration, as well as the fps and mks unit systems. 1. Integration of y':2x+l yields y(x) = I(2x+1)dx = x2+x+C. Then substitution of x20. 31:3 gives 3 = 0+0+C = C, so y(x) : x2+x+3. 2. Integration of y'=(;t—2)2 yields y(x) = I(x-—2)2dx 2 §(x—2)3+C. Then substitution of x=2, y=l gives 1 = 0+C = C, so y(x) = ﬂat—2)“. 3. Integration of y’:\/; yields y(x) = Jﬁdx 2 %xllf2+C- Then substitution of 1:4, y=0 gives 0=%+C, SO y(x) = %(X312*8)_ 4. IntegratiOn of y'zx'2 yields y(x) = [:6de = —1/x+C. Then substitution of x:], y=5 gives 5=—1+C, so y(x) = —]/x+6. 5. Integration of y' 2 (x + 2T1” yields y(x) = J(x+ 2)";2 dx = 2dx+2 +C. Then substitution of x = 2, y z —1 gives ~l = 2-2+ C, so y(x) = 24x+ 2 —5. 6. Integration of y’=x(x-2+9)”2 yields y(x) = Ix(x2+9)mdx = §(x3+9)m+c. Then substitution of x=-—4, y:0 gives 0=_%(5)3+C, so y(x) = {of +9)“ 425]. 7. Integration of y’:'10f(x2+1) yields y(x) 2 IiO/(x2+l)dx = lOtan"x+C. Then substitution of x=0, y=0 gives 0:10-0+C, so y(x) = lOtan_'x. 8. Integration of y’zcost yields 390:) = Icoshdx = 35in 2x+C. Then substitution of x=0, y=1 gives 1=0+C, so y(x) = ésin2x+l. 9. Integration of y'=1f~Jl—Ax2 yields y(x) : Illdl—xzdx = sin"'x+C. Then substitution of x: 0, y :0 gives 020+ C, so y{x) = sin“‘x. 10. Integratibn of y’zxe" yields y”) : ixeﬁdx : .iueﬂd“ = (““UE" = ”(x+l)c'I+C Section 1 .2 5 ...
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