Question

In Problems 21-26 solve the given initial-value problem

$\left(\frac{3y^2-t^2}{y^5}\right)\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{t}}{\mathbf{2}{\mathbf{y}}^{\mathbf{4}}}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The initial value of the problem is$\frac{t^2}{4y^4}-\frac{3}{2y^2}=-\frac{5}{4}$

Step-by-step solution

Given Information

The given equation is,

$\begin{array}{l}\left(\frac{3y^2-t^2}{y^5}\right)\frac{dy}{dt}+\frac{t}{2y^4}=0\\ \frac{t}{2y^4}dt+\frac{3y^2-t^2}{y^5}dy=0\end{array}$

Compare the given equation with $Mdx+Ndy=0$

$\begin{array}{l}M(t,y)=\frac{t}{2y^4}\\ N(t,y)=\frac{3y^2-t^2}{y^5}\end{array}$

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial t}$

$\begin{array}{l}\frac{\partial M}{\partial y}=\frac{-2t}{y^5}\\ \frac{\partial N}{\partial t}=-\frac{2t}{y^5}\end{array}$

Since, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$

Therefore, the equation is exact.

Find the solution

Now, consider the equation as, $f(t,y)=\frac{t^2}{4y^4}+g\left(y\right)$

Integrate $M(t,y)=\frac{\partial f}{\partial t}=\frac{t}{2y^4}$

$\frac{\partial f}{\partial y}=-\frac{t^2}{y^5}+g^{¢}\left(y\right)$

$g^{¢}\left(y\right)=\frac{3y^2}{y^5}=\frac{3}{y^3}$

Since$N(t,y)=\frac{\partial f}{\partial y},$we see that

$\begin{array}{l}g\left(y\right)=-\frac{3}{2y^2}\\ g^{¢}\left(y\right)=\frac{3y^2}{y^5}\end{array}$

So, the solution is $\frac{t^2}{4y^4}-\frac{3}{2y^2}=c$

Substitute the initial condition $y\left(1\right)=1$

$\begin{array}{c}\frac{1}{4}-\frac{3}{2}=c\\ c=-\frac{5}{4}\end{array}$

Hence, the required solution is,

$\frac{t^2}{4y^4}-\frac{3}{2y^2}=-\frac{5}{4}$