Question

Question: In Problems 1-30, solve the equation.

${\mathbf{t}}^3{\mathbf{y}}^2\mathbf{dt}\mathbf{+}{\mathbf{t}}^4{\mathbf{y}}^{-6}\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

$\mathbf{t}\mathbf{=}{\mathbf{Ce}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^7}\right)}$

Step-by-step solution

Definition and concepts to be used

Definition of Initial Value Problem:By an initial value problem for an nth-order differential equation $\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}\mathrm{...}\mathbf{,}\frac{{\mathbf{d}}^n\mathbf{y}}{{\mathbf{dx}}^n}\right)\mathbf{=}\mathbf{0}$ we mean: Find a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_0\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_1\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\mathbf{,}\end{array}$

Where ${\mathbf{x}}_0\in \mathbf{I}$ and ${\mathbf{y}}_0\mathbf{,}{\mathbf{y}}_1\mathbf{,}\mathrm{...}\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Formulae to be used:

• Integration by parts:$\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$.
• $\int {\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}a$.

Given information and simplification

Given that,${\mathbf{t}}^3{\mathbf{y}}^2\mathbf{dt}\mathbf{+}{\mathbf{t}}^4{\mathbf{y}}^{-6}\mathbf{dy}\mathbf{=}\mathbf{0}\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluate the equation (1).

$\begin{array}{c}{\mathbf{t}}^3{\mathbf{y}}^2\mathbf{dt}\mathbf{+}{\mathbf{t}}^4{\mathbf{y}}^{-6}\mathbf{dy}\mathbf{=}\mathbf{0}\\ {\mathbf{t}}^3{\mathbf{y}}^2\mathbf{dt}\mathbf{=}\mathbf{-}{\mathbf{t}}^4{\mathbf{y}}^{-6}\mathbf{dy}\\ \frac{{\mathbf{t}}^3}{{\mathbf{t}}^4}\mathbf{dt}\mathbf{=}\mathbf{-}\frac{{\mathbf{y}}^{-6}}{{\mathbf{y}}^2}\mathbf{dy}\\ \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-8}\mathbf{dy}\mathrm{......}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Now integrate the equation (2) on both sides.

$\int \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}\int {\mathbf{y}}^{-8}\mathbf{dy}\mathrm{......}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Evaluation method

$\begin{array}{c}\int \frac{1}{t}\mathbf{dt}\mathbf{=}\mathbf{-}\int {\mathbf{y}}^{-8}\mathbf{dy}\\ \mathbf{ln}\left|t\right|\mathbf{+}{\mathbf{C}}_1\mathbf{=}\frac{1}{7}{\mathbf{y}}^{-7}\\ \mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\mathbf{=}{\mathbf{y}}^{-7}\\ \mathbf{y}\mathbf{=}{\left(\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\right)}^{\mathbf{-}\frac{1}{7}}\end{array}$

Take 7^th root on both sides.

$\begin{array}{c}{\mathbf{y}}^7\mathbf{=}\frac{1}{\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}}\\ {\mathbf{y}}^7\left(\mathbf{7}\mathbf{ln}\left|t\right|\mathbf{+}\mathbf{C}\right)\mathbf{=}\mathbf{1}\\ \mathbf{t}\mathbf{=}{\mathbf{Ce}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^7}\right)}\end{array}$

Hence, the solution of the given initial value problem is $\mathbf{t}\mathbf{=}{\mathbf{Ce}}^{\left(\frac{1}{\mathbf{7}{\mathbf{y}}^7}\right)}$.