Question

In Problems, 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\mathbf{(}\mathbf{4}{\mathit{t}}^{\mathbf{3}}\mathit{y}\mathbf{-}\mathbf{15}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{y}\mathbf{)}\mathit{d}\mathit{t}\mathbf{+}\mathbf{(}{\mathit{t}}^{\mathbf{4}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathit{t}\mathbf{)}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The answer is$c=t^{4}y-5t^3+y^3-ty$

Step-by-step solution

Given Information.

The given equation is$(4t^{3}y-15t^2-y)dt+(t^4+3y^2-t)dy=0$

Setting M and N values

For the given equation, M should be equals to $(t,y)dt$and N should be equals to$(t,y)dy$.

$$\begin{gathered} M=(4t^{{}^3}y-15t^2-y)dt \\ N=(t^4+3y^2-t)dy \end{gathered}$$

 Determining partial derivatives

M with respect to y and N with respect to t are partial derivatives that must be evaluated

$\frac{\partial M}{\partial y}=4t^3-1$

$\frac{\partial N}{\partial t}=4t^3-1$

As a result, the equation is exact. .$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$

Calculate M and N's integrals

Now, find the integrals of M and N.

$\begin{array}{c}M=\int (4t^{3}y-15t^2-y)dt\\ =t^{4}y-5t^3-ty+g\left(y\right)\end{array}$

$\begin{array}{c}N=\int (t^4+3y^2-t)dy\\ =t^{4}y+y^3-ty+h\left(t\right)\end{array}$

Equate the resulting expressions to c, making a note of any recurring terms.

$c=t^{4}y-5t^3+y^3-ty$

So, the result is $c=t^{4}y-5t^3+y^3-ty$