Question

Solve the following differential equation,

(2 t x + 3 t^2) x′ = x^2

with initial conditions

(a) x(1) = 5
(b) x(1) = −3

write answers a and b as a function of t

Answer 1

Explanation:

To solve the differential equation (2tx + 3t^2)x' = x^2, we can use separation of variables method:

First, we separate the variables:

x^-2 dx/dt = (2tx + 3t^2)^-1 dt

Next, we integrate both sides:

∫x^-2 dx = ∫(2tx + 3t^2)^-1 dt

Using u-substitution with u = 2tx + 3t^2, du/dt = 2x + 6t, and du = (2x + 6t)dt, we get:

-x^-1 = (1/6) ln|2tx + 3t^2| + C

where C is the constant of integration.

Now, we can use the initial conditions to solve for the constant C:

(a) x(1) = 5:

-5^-1 = (1/6) ln|2t(1) + 3(1)^2| + C

-6 = ln|2t + 3| + C

C = -6 - ln|2t + 3|

Therefore, the solution for x as a function of t with initial condition x(1) = 5 is:

x(t) = (-1/6) - (1/6) ln|2t + 3|

(b) x(1) = −3:

-3^-1 = (1/6) ln|2t(1) + 3(1)^2| + C

-2 = ln|2t + 3| + C

C = -2 - ln|2t + 3|

Therefore, the solution for x as a function of t with initial condition x(1) = −3 is:

x(t) = (-1/3) - (1/6) ln|2t + 3|