# Solve the differential equation below with the given initial condition of P(1)=2 and find the...

## Question:

Solve the differential equation below with the given initial condition of {eq}P(1)=2 {/eq} and find the function {eq}P(t) {/eq}.

{eq}\frac{dP}{dt}=e^{-P} \sqrt t {/eq}

## Differential Equation:

The differential equation in which the variables can be separated is called separable differential equation.

It can be solved using general antiderivatives.

Formulas Used:

1.{eq}\displaystyle \int e^{ax}dx=\frac{e^{ax}}{a}+c {/eq}.

2.{eq}\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+c {/eq}

## Answer and Explanation:

Given equation {eq}\displaystyle \frac{dP}{dt}=e^{-P} \sqrt t {/eq} and {eq}\displaystyle P(1)=2 {/eq}.

Now,

{eq}\displaystyle \begin{align} \frac{dP}{dt}&=e^{-P} \sqrt t\\ e^{P}dP&=\sqrt t \ dt\\ \int e^{P}dP&=\int \sqrt t \ dt\\ e^P&=\frac{2}{3}t\sqrt t+c\\ e^{P(1)}&=\frac{2}{3}+c\\ c&=e^2-\frac{2}{3}. \end{align} {/eq}

{eq}\displaystyle \Rightarrow {/eq}The equation is {eq}\displaystyle e^P=\frac{2}{3}t\sqrt t+e^2-\frac{2}{3} {/eq}

{eq}\displaystyle \Rightarrow P=\ln\left( \frac{2}{3}t\sqrt t+e^2-\frac{2}{3} \right) {/eq}

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from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 1