Solve the differential equation
#dy/dt = 4 sqrt(yt)# y(1)=6?

Question

Solve the differential equation 
#dy/dt = 4 sqrt(yt)# y(1)=6?

Emilia Aguilar · Accepted Answer

# y = 16/9 ( t^(3/2) + (3 sqrt(6))/4 - 1)^2#

and:

# y = 16/9 ( t^(3/2) - (3 sqrt(6))/4 - 1)^2# #dy/dt = 4 sqrt(yt)# This is separable. #1/sqrt y dy/dt = 4 sqrt t# Differentiate both sides wrt t: #int 1/sqrt y dy/dt \dt = 4 int sqrt t \ dt# Chain rules allows us to re-write first term: #int 1/sqrt y \ dy = 4 int sqrt t \ dt# Then integrate: #2 sqrt y = 4 2/3 t^(3/2) + C# #implies sqrt y = 4/3 t^(3/2) + C# #implies y = 16/9 ( t^(3/2) + C)^2# Apply the IV: #y(1) = 6# #implies 6 = 16/9 ( 1 + C)^2# #implies C = pm sqrt(54/16) - 1 = pm (3 sqrt(6))/4 - 1# So: #implies y = 16/9 ( t^(3/2) + (3 sqrt(6))/4 - 1)^2# And: #implies y = 16/9 ( t^(3/2) - (3 sqrt(6))/4 - 1)^2#

Evelyn Agard · Answer

#y=(4/3t^(3/2)+sqrt6-4/3)^2#

We should separate the variables here by treating #dy/dt# as a division problem to get all terms with #y# and all terms with #t# on the same side of the equation.

#dy/dt=4sqrtysqrtt#

#dy/sqrty=4sqrttdt#

Integrating both sides and rewriting with fractional exponents:

#inty^(-1/2)dy=4intt^(1/2)dt#

Using typical integration rules:

#y^(1/2)/(1/2)=4(t^(3/2)/(3/2))+C#

#2sqrty=8/3t^(3/2)+C#

Solving for #y# gives:

#y=(4/3t^(3/2)+C)^2#

We were given the initial condition #y(1)=6#, which we can use to solve for #C#:

#6=(4/3(1)^(3/2)+C)^2#

#sqrt6=4/3+C#

#C=sqrt6-4/3#

Then:

#y=(4/3t^(3/2)+sqrt6-4/3)^2#

Luke Alvarez · Answer

undefined The solution to the given differential equation $ \frac{dy}{dt} = 4\sqrt{yt} $ with the initial condition $ y(1) = 6 $ is $ y(t) = (t^2 + 1)^2 $.

Solve the differential equation #dy/dt = 4 sqrt(yt)# y(1)=6?