# Suppose that y' = k y, y(0) = 3, and y'(0) = 7. What is y as a function of t?

## Question:

Suppose that {eq}\displaystyle y' = k y, y(0) = 3, \ and \ y'(0) = 7.{/eq} What is {eq}y {/eq} as a function of {eq}t? {/eq}

## Differential Equations:

The equation formed by the differentials is called a differential equation.

A separable differential equation can be solved using anti derivatives.

Formulas Used:

1.{eq}\displaystyle (\ln x)'=\frac{1}{x} {/eq}.

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

Given function {eq}\displaystyle y'=ky {/eq} and {eq}\displaystyle y(0) = 3, \ y'(0) = 7. {/eq}

Now,

{eq}\displaystyle \begin{align} y'(0)&=ky(0)\\ 7&=3k\\ k&=\frac{7}{3}. \end{align} {/eq}

The equation is {eq}\displaystyle y'=\frac{7}{3}y {/eq}.

Since {eq}\displaystyle y {/eq} is a function of {eq}\displaystyle t {/eq}.

{eq}\displaystyle \begin{align} \frac{dy}{dt}&=\frac{7}{3}y\\ \frac{dy}{y}&=\frac{7}{3}dt\\ d(\ln y)&=\frac{7}{3}dt\\ \int d(\ln y)&=\frac{7}{3}\int dt\\ \ln y&=\frac{7t}{3}+c\\ \ln y(0)&=c\\ c&=\ln 3. \end{align} {/eq}

{eq}\displaystyle \Rightarrow {/eq}The function is {eq}\displaystyle \ln y=\frac{7t}{3}+\ln 3 {/eq}.

{eq}\displaystyle \Rightarrow y(t)= \displaystyle e^{ \left(\displaystyle \frac{7t}{3}+ \displaystyle \ln 3\right)} {/eq}