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m_a_m_a [10]
4 years ago
15

Solve the differential equation t dy/dt + dy/dt = te ^y

Mathematics
1 answer:
LuckyWell [14K]4 years ago
5 0

The answer is

t dy / dt + dy / dt = te ^ y

I apply common factor                1/dt*(tdy+dy)=te^y

I pass "dt"                    tdy+dy=(te^y)*dt

I apply common factor                (t+1)*dy=(te^y)*dt

I pass "e^y"                         (1/e^y)dy=((t+1)*t)*dt

I apply integrals                ∫ (1/e^y)dy= ∫ (t^2+t)*dt

by property of integrals  ∫ (1/e^y)dy= ∫ (e^-y)dy

∫ (e^-y)dy= ∫ (t^2+t)*dt

I apply integrals

-e^-y=(t^3/3)+(t^2/2)+C

I apply natural logarithm to eliminate "e"

-ln (e^-y)=-ln(t^3/3)+(t^2/2)+C

y=ln(t^3/3)+(t^2/2)+C

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
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\end{array}

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% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\end{array}

\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
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now, with that template in mind, let's take a peek at yours

\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
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