Question

Question: In Problems $\mathbf{17}\mathbf{-}\mathbf{22}$, solve the initial value problem.

${\mathbf{t}}^{\mathbf{2}}\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{3}\mathbf{tx}\mathbf{=}{\mathbf{t}}^{\mathbf{4}}\mathbf{lnt}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{x}\left(1\right)\mathbf{=}\mathbf{0}$

Short answer

The initial value to the given equation is$\mathrm{x}=\frac{{\mathrm{t}}^3}{6}\mathrm{lnt}-\frac{{\mathrm{t}}^3}{36}+\frac{1}{2\mathrm{t}}-\frac{17}{36{\mathrm{t}}^3}$.

Step-by-step solution

Method for solving linear equations

• $\frac{\mathbf{d}}{\mathbf{dx}}\left[\mu \left(x\right)y\right]$Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$by the formula $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{e}\mathit{x}\mathit{p}\left[{\int }_{}^{}P\left(x\right)\mathrm{dx}\right]$.
• Multiply the equation in standard form by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by $\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$. Here C is an arbitrary constant.

Solve the given equation

Given that,

${\mathrm{t}}^2\frac{\mathrm{dx}}{\mathrm{dt}}+3\mathrm{tx}={\mathrm{t}}^4\mathrm{lnt}+1,\mathrm{x}\left(1\right)=0$

Write the equation (1) into standard form of linear equation.

$$\begin{gathered} {\mathrm{t}}^2\frac{\mathrm{dx}}{\mathrm{dt}}+3\mathrm{tx}={\mathrm{t}}^4\mathrm{lnt}+1 \\ \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{3\mathrm{x}}{\mathrm{t}}={\mathrm{t}}^2\mathrm{lnt}+\frac{1}{{\mathrm{t}}^2}.........\left(2\right) \end{gathered}$$

Calculate the integrating factor of $\mu \left(t\right)$.

Where $\mathrm{P}\left(\mathrm{t}\right)=\frac{3}{\mathrm{t}}$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}\frac{3}{\mathrm{t}}\mathrm{dt}}\\ & =& {\mathrm{e}}^{3\mathrm{lnt}}\\ & =& {\mathrm{t}}^3\end{array}$

Simplification method

Multiply $\mathrm{\mu }\left(\mathrm{t}\right)$ in equation (2)

$$\begin{gathered} {\mathrm{t}}^3\frac{\mathrm{dx}}{\mathrm{dt}}+{\mathrm{t}}^3\frac{3\mathrm{x}}{\mathrm{t}}={\mathrm{t}}^3{\mathrm{t}}^2\mathrm{lnt}+{\mathrm{t}}^3\frac{1}{{\mathrm{t}}^2} \\ {\mathrm{t}}^3\frac{\mathrm{dx}}{\mathrm{dt}}+3{\mathrm{xt}}^2={\mathrm{t}}^5\mathrm{lnt}+\mathrm{t} \\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{t}}^3\right]\mathrm{x}={\mathrm{t}}^5\mathrm{lnt}+\mathrm{t} \end{gathered}$$

Integrating both sides.

role="math" localid="1664162802857" $$\begin{gathered} {\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{t}}^3\mathrm{x}\right]\mathrm{xdt}={\int }_{}^{}\left({\mathrm{t}}^5\mathrm{lnt}+\mathrm{t}\right)\mathrm{dt} \\ {\mathrm{t}}^3\mathrm{x}={\int }_{}^{}{\mathrm{t}}^5\mathrm{lnt}\mathrm{dt}+\frac{1}{2}{\mathrm{t}}^2+\mathrm{C}.......\left(3\right) \end{gathered}$$

Find the value of ${\int }_{}^{}{\mathrm{t}}^5\mathrm{lnt}\mathrm{dt}$separately.

Let $\mathrm{u}=\mathrm{lnt},\mathrm{v}={\mathrm{t}}^5\mathrm{dt}$

Then,

$\begin{array}{rcl}{\int }_{}^{}{\mathrm{t}}^5\mathrm{lnt}\mathrm{dt}& =& \frac{{\mathrm{t}}^6\mathrm{lnt}}{6}-{\int }_{}^{}\frac{{\mathrm{t}}^5}{6}\\ & =& \frac{{\mathrm{t}}^6\mathrm{lnt}}{6}-\frac{{\mathrm{t}}^6}{36}\end{array}$

Substitute the value in equation (3).

$$\begin{gathered} {\mathrm{t}}^3\mathrm{x}=\frac{{\mathrm{t}}^6\mathrm{lnt}}{6}-\frac{{\mathrm{t}}^6}{36}+\frac{1}{2}{\mathrm{t}}^2+\mathrm{C} \\ \mathrm{x}=\frac{{\mathrm{t}}^3\mathrm{lnt}}{6}-\frac{{\mathrm{t}}^3}{36}+\frac{1}{2\mathrm{t}}+\frac{\mathrm{C}}{{\mathrm{t}}^3}.....\left(4\right) \end{gathered}$$

Find the initial value

Given that, $x\left(1\right)=0$.

Then, x = 0 and t = 1.

Substitute the value in equation (4) to get the value of C.

$$\begin{gathered} \mathrm{x}=\frac{{\mathrm{t}}^3\mathrm{lnt}}{6}-\frac{{\mathrm{t}}^3}{36}+\frac{1}{2\mathrm{t}}+\frac{\mathrm{C}}{{\mathrm{t}}^3} \\ 0=\frac{\ln \left(1\right)}{6}-\frac{1}{36}+\frac{1}{2}+\mathrm{C} \\ \mathrm{C}=\frac{1}{36}-\frac{1}{2} \\ \mathrm{C}=-\frac{17}{36} \end{gathered}$$

Substitute the value of C in equation (2).

$\mathrm{x}=\frac{{\mathrm{t}}^3\mathrm{lnt}}{6}-\frac{{\mathrm{t}}^3}{36}+\frac{1}{2\mathrm{t}}-\frac{17}{36{\mathrm{t}}^3}$

So, the solution is $\mathbf{x}\mathbf{=}\frac{{\mathbf{t}}^{\mathbf{3}}\mathbf{lnt}}{\mathbf{6}}\mathbf{-}\frac{{\mathbf{t}}^{\mathbf{3}}}{\mathbf{36}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}\mathbf{t}}\mathbf{-}\frac{\mathbf{17}\mathbf{}}{\mathbf{36}{\mathbf{t}}^{\mathbf{3}}}$