Let's rewrite the function: 2ty '= 4y 2t (dy / dt) = 4y t (dy / dt) = 2y We use the separable variables method: (dy / y) = 2 (dt / t) We integrate both sides: ln (y) = 2ln (t) + C we apply exponential to both sides: exp (ln (y)) = exp (2ln (t) + C) Exponential properties exp (ln (y)) = exp (2ln (t)) * exp (C) Log properties: exp (ln (y)) = Cexp (ln (t ^ 2)) Exponential properties: y = C * t ^ 2 Initial conditions y (2) = - 8: -8 = C * (2) ^ 2 We clear C: C = -8 / (2 ^ 2) = - 8/4 = -2 The function is: y = -2 * t ^ 2 Therefore we have to compare: y = -2 * t ^ 2 y = ct ^ r The values of c and r are: c = -2 r = 2 Answer: the value of the constant c and the exponent r are: c = -2 r = 2 so that y = -2 * t ^ 2 is the solution of this initial value problem
2ty'=4y Replacing y'=dy/dt in the equation: 2t(dy/dt)=4y Grouping terms: dy/y=4dt/(2t) dy/y=2dt/t Integrating both sides: ln(y)=2ln(t)+ln(c), where c is a constant Using property logarithm: b ln(a) = ln(a^b), with b=2 and a=t ln(y)=ln(t^2)+ln(c) Using property of logarithm: ln(a)+ln(b) = ln(ab), with a=t^2 and b=c ln(y)=ln(ct^2) Then: y=ct^2 Using the initial condition: y(2)=-8 t=2→y=-8→c(2)^2=-8→c(4)=-8 Solving for c: c=-8/4 c=-2
Then the solution is y=-2t^2 Comparing with the solution: y=ct^r c=-2, r=2
Answer: T<span>he value of the constant c is -2 (c=-2) and the exponent r is 2 (r=2)</span>
