# y ^1 (t) = \frac{2 t ^2 + 4}{t}, y(0) = 2

## Question:

{eq}y ^1 (t) = \frac{2 t ^2 + 4}{t}, y(0) = 2 {/eq}

## Initial Value Problem in Calculus:

An initial value problem is also called a Cauchy problem. We are given a separable ordinary differential equation of first order and first degree {eq}\frac{dy}{dt}=f\left( t,y\right) {/eq}.

To solve this problem, we'll integrate both sides. Next, we'll plug in the given initial condition to get the value of the constant. In this problem, the value of the constant is undefined.

We are given:

{eq}y ^1 (t) = \frac{2 t ^2 + 4}{t} {/eq}

Rewrite:

{eq}\Rightarrow\displaystyle \dfrac{\ dy }{ \ dt } = \frac{2 t ^2 + 4}{t} {/eq}

{eq}\Rightarrow\displaystyle \ dy = \frac{2 t ^2 + 4}{t} \ dt {/eq}

Integrate both sides:

{eq}\Rightarrow\displaystyle \int \ dy = \int \frac{2 t ^2 + 4}{t} \ dt {/eq}

Now solve {eq}\displaystyle \int \ dy = y+c_{1} {/eq}

Next solve {eq}\displaystyle \int \frac{2 t ^2 + 4}{t} \ dt {/eq}

{eq}= \displaystyle \int \left( \frac{2 t ^2 }{t}+\dfrac {4}{t} \right)\ dt {/eq}

Apply integral power rule:

{eq}= \displaystyle \int \frac{2 t ^2 }{t} \ dt + \int \dfrac {4}{t} \ dt {/eq}

Take the constant out:

{eq}= \displaystyle 2 \int t\ dt +4 \int \dfrac {1}{t} \ dt {/eq}

{eq}= \displaystyle 2 \dfrac{t^2}{2}+4 \ln|t|+c_{2} {/eq}

{eq}= \displaystyle t^2+4 \ln|t|+c_{2} {/eq}

Plug in back and combine the constant:

{eq}\Rightarrow \displaystyle y= t^2+4 \ln|t|+C {/eq}

Plug in the given condition {eq}y(0)=2 {/eq}

{eq}\displaystyle 2= 0^2+4 \ln(0)+C {/eq}

Now we know that {eq}\ln(0) {/eq} undefined. So the value of the constant is undefined.

Therefore the solution is: {eq}\displaystyle {\boxed{ y (t) = t^2+4 \ln|t|+ undefined }} {/eq}

Differential Calculus: Definition & Applications

from Calculus: Help and Review

Chapter 13 / Lesson 6
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