find-the-solution-to-the-initial-value-problem-t4dzdtz29displaystyle-t-4-fracdzd

Question

Find the solution to the initial value problem: (t+4)dzdt=z2+9\displaystyle (t + 4) \frac{dz}{dt} = $$z^2 + 9  $$for  -4 "\u003et\u003e−4 t \u003e -4 , z(0)=0 z(0) = 0 .

Accepted Answer

z=3tan⁡(3ln⁡(t+44))\displaystyle z = 3 	an\left(3\ln\left(\frac{t + 4}{4}ight)ight)

Answer

z=3tan⁡(ln⁡(t+44)){\displaystyle z=3	an\left(\ln\left(\frac{t + 4}{4}ight)ight)}

Answer

z=3tan⁡(3ln⁡(t4)){\displaystyle z=3	an\left(3\ln\left(\frac{t}{4}ight)ight)}

Answer

z=tan⁡(3ln⁡(t+44)){\displaystyle z=	an\left(3\ln\left(\frac{t + 4}{4}ight)ight)}

Find the solution to the initial value problem: (t+4)dzdt=z2+9\(\displaystyle\) (t + 4) \(\frac{dz}{dt}\) = z^2 + 9 for t>−4 t > -4 , z(0)=0 z(0) = 0 .

Find the solution to the initial value problem: $(t + 4) \frac{d z}{d t} = z^{2} + 9$ for $t is greater than negative 4$ , $z of 0 equals 0$ .