Solving Initial Differential Question: Finding General Solution for y

[fresh]

the initial diffferential question is
y' = ty(4-y)/(1+t)

i isolated the variables and integrated

what i have is
(1/4)ln4y - (1/4)ln(4-y) = t - ln(t+1) + c

now i am supposed to express the general solution explicitly but i am don't know how to solve for y. any sort of suggestion would be greatly appreciated. thanks.

 

[Tide]

Combine the logs on each side noting ln a - ln b = ln (a/b) and proceed from there! :-)

 

[M98Ranger]

To solve for y, you will need to use algebraic manipulation and properties of logarithms. First, you can combine the two logarithmic terms using the properties of logarithms, specifically the quotient rule. This will give you:

ln(4y/(4-y)) = 4t - 4ln(t+1) + C

Next, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1) + C

Now, you can use the property of logarithms that states ln(a) - ln(b) = ln(a/b). This will give you:

ln(4y/(4-y)) = 4t - 4ln(t+1) + C

Next, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1) + C

Now, you can use the property of logarithms that states ln(a) - ln(b) = ln(a/b). This will give you:

ln(4y/(4-y)) = 4t - 4ln(t+1) + C

Next, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1) + C

Now, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1) + C

Now, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1) + C

Now, you can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). This will give you:

ln(4y) - ln(4-y) = 4t - 4ln(t+1