Solve the initial problem: dy/dx = y/(x^2-4) y(0)=2 Solve the initial problem: d

Question

Solve the initial problem: dy/dx = y/(x^2-4) y(0)=2 Solve the initial problem: dy/dx = (2y+5)/(x+8) y(-3)=35 Solve the initial problem: dy/dx = x(y^2+9)/y(x^2+4) y(1)=6

Explanation / Answer

dy/dx = y/(x^2-4) , y(0)=2 => (1/y) dy = 1/(x^2 - 4) dx => ln y = (1/4)* ln |(x-2)/(x+2)| + C put y(0)=2, => ln 2 = (1/4)*ln |-2/2| + C => ln 2 = (1/4)*ln 1 + C => ln 2 = C therefore, ln y = (1/4)*ln |(x-2)/(x+2)| + ln 2 =>ln (y/2) = ln {|(x-2)/(x+2)|^(1/4)} => y/2 = |(x-2)/(x+2)|^(1/4) => y = 2 * {|(x-2)/(x+2)|^(1/4) } => y^4 = 16 * |(x-2)/(x+2)| answer.... dy/dx = (2y+5)/(x+8) , y(-3)=35 =>1/(2y+5) dy = 1/(x+8) dx => (1/2)*ln|2y+5| = ln|x+8| + C put y(-3) = 35 therefore, => (1/2)*ln 1 = ln 5 + C => C = - ln 5 therefore, (1/2)*ln|2y+5| = ln|x+8| - ln 5 => 5*(2y+5)^(1/2) = (x+8) => 25*(2y+5) = (x+8)^2 ....Answer.... dy/dx = x(y^2+9)/y(x^2+4) , y(1)=6 =>y/(y^2+9) dy = x/(x^2+4) dx =>(1/2) ln (y^2+9) = (1/2) ln (x^2+4) + C put y(1) = 6 therefore, =>(1/2) ln 45 = (1/2) ln 5 + C =>(1/2)[ln 45 - ln 5] = C =>(1/2) ln 9 = C => ln 3 = C therefore, (1/2) ln (y^2+9) = (1/2) ln (x^2+4) + ln 3 => ln (y^2+9) = (1/2) ln (x^2+4) + ln 9 => y^2+9 = (x^2+4)*9 => y^2+9 = 9*x^2+36 => y^2 = 9*x^2+27 Answer.......

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