Solve the following ordinary differential equation by the method of separation o

Question

Solve the following ordinary differential equation by the method of separation of variables. dz/dt = -2(5 - z3)/tz2 when z(1) = 0 Recall that some differential equations can be written in the form q(z) z' = p(t) When this happens then one can write q(z) dz = p(t)dt + C and have an implicit solution. In some cases this the implicit solution can be arranged to find an explicit solution z(t). What is q(z) in this case? q(z) = Enter just an expression for q(z). Do not include z' What is p(t) in this case? p(t) = Enter just an expression for p(t). Calculate an antiderivative of q(z)? q(z) dz = Enter just an expression in z. Do not include a constant of integration! Calculate an antiderivative of p(t)? p(t) dt = Enter just an expression in t. Do not include a constant of integration! What is the value of the constant C in the equation q(z) dz = p(t) dt + C? C = Calculate an exact number C using the given information, e.g. ln(8)/3. Do not give a decimal answer. What is the explicit solution of the given differential equation? Enter just an equation of the form z = (some expression in t only).

Explanation / Answer

so we move anything with z over and we get z^2/ (5-z^3) z' = -2/t 1) q=z^2/ (5-z^3) 2) p=-2/t 3) integral of z^2/(5-z^3) let u = 5-z^3 then du= -3z^2 dz so this becomes integral of -1/3 du/u = -1/3 ln u = -1/3 ln(5-z^3) 4) integral of -2/t = -2 ln t 5) so we have -1/3 ln(5-z^3) = -2 lnt + C we know z(1)=0 so we have -1/3 ln(5) = -2 ln 1 + C C= -1/3 ln(5) so we have -1/3 ln(5-z^3)= -2 ln t - 1/3 ln (5) ln (5-z^3) =6 ln t + ln(5) = ln t^6 + ln(5) raise both sides to the e 5-z^3 = e^(lnt^6 + ln(5) = e^ lnt^6 * e^ ln(5) 5-z^3 = 5 t^6 z^3 = 5-5t^6 z= (5-5t^6)^(1/3)

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