Introduction : Here we shall solve dy/dt=yt^3-1.5y subject to y(0)=1 which is a ordinary differential equation of first order with a condition on solution.
Question : Solve
\[\frac{dy}{dt}=yt^3-1.5y , y(0)=1\]
Solution :
Given differential equation
\[\frac{dy}{dt}={\rm yt}^3-1.5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \]
Subject to
\[\ y\left(0\right)=1\]
Eqn. (1) can be written as
\[\frac{dy}{dt}=\left(t^3-1.5\right)y\]
This is a separable form of an ode which can be written in separable form as follows
\[\frac{1}{y}dy=\left(t^3-1.5\right)dt\]
Integrating both sides
\[\log{\left(y\right)}=\frac{t^4}{4}-1.5t+c\]
Where c is integrating constant. Taking antilogarithm both sides, we obtained
\[y\left(t\right)=e^{\frac{t^4}{4}-1.5t+c}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\]
This is the general solution of ODE (1).
Now to determine particular solution of eqn. (1) we shall substitute t =0 in eqn (2)
to get
\[y\left(0\right)=1\] \[y\left(0\right)=e^{0+c}\]
Now given condition is that
\[ y\left(0\right)=1.
\[1=e^c\]
\[c=\log{\left(1\right)}=0\]
We shall substitute value of c into eqn (2) to obtain
\[y\left(t\right)=e^{\frac{t^4}{4}-1.5t}\]
This is the required solution of eqn. (1) that satisfies the given condition.