Find the particular solution of the differential equation (x ^2) /(y ^2 ?(3) dy

Question

Find the particular solution of the differential equation (x ^2) /(y ^2 ?(3) dy /dx =1 2y satisfying the initial condition y(1)=sqrt(4) Answer: y=

Explanation / Answer

uk.answers.yahoo.com/question/index?qid=20110116085220... see it or this type of DE is classified as separable, i.e. dy/dx = F(x)G(y) Thus, dy/G(y) = F(x)dx and so you integrate both sides and solve for y. here dy/dx = y^2 - y - 2 dy / (y^2 - y - 2 ) = dx integrate both sides int ( 1/(y^2 - y - 2 ) dy ) = int (1 dx ) = x + C LHS = int ( 1/ (y^2 -y - 2) dy ) = int ( 1 / ( (y - 2)(y + 1) ) dy ) Decompose into Partial Fractions 1 / ( (y - 2)(y + 1) ) = k1 / ( y - 2 ) + k2 / ( y + 1) 1 = k1(y + 1) + k2(y - 2) (which holds for all y) let y = -1 1 = -3k2 --> k2 = -1/3 let y = 2 1 = 3k1 --> k1 = 1/3 Thus, 1 / ( (y - 2)(y + 1) ) = (1 / 3) / ( y - 2 ) + (-1 / 3) / ( y + 1) = (1/3) ( 1 / (y - 2) - 1 /(y + 1) ) Hence our integral becomes, int ( 1 / ( y - 2)( y + 1) dy ) = (1/3) * [ int ( 1 / (y - 2) dy ) - int ( 1 / (y + 1) dy ) ] = (1/3) [ ln (y -2) - ln(y + 1) ] = (1/3) ln ( (y - 2) / (y + 1) ] so equating LHS = RHS we have (1/3) ln( (y - 2) / (y + 1) ) = x + C ln ( (y - 2) / (y + 1) ) = 3x + D ( D = 3C ) (y - 2)/(y + 1) = exp(3x + D) y - 2 = (y + 1)exp(3x + D) y - 2 = y * exp(3x + D) + exp(3x + D) y(1 - exp(3x + D) ) = exp(3x + D) + 2 y = ( exp(3x + D) + 2 ) / ( 1 - exp(3x + D) ) Hope this helps, plz rate me

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