solve dy/dt=yt^3-1.5y subject to y(0)=1

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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.


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