Consider the differential equation dxdt-10-0-2x with initial condition x0 - 1- The response xt for t - 0 is

Dxdt=10 0.2x dx10 0.2x=dt Intergrating both sides log(10 0.2x)( 0.2)=t +C x(0) = 1 ⇒ C = log9.8( 0.2) log(10 0.2x)( 0.2) log9.8( 0.2)=t log(10 0.2x)9.8= ...

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Consider the ...

Question

Consider the differential equation $\frac{d x}{d t} = 10 - 0.2 x$ with initial condition x(0) = 1, The response x(t) for t > 0 is

2 - e^{- 0.2 t}

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2 - e^{0.2 t}

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50 - 49 e^{- 0.2 t}

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50 - 49 e^{0.2 t}

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Solution

The correct option is C $50 - 49 e^{- 0.2 t}$
$\frac{d x}{d t} = 10 - 0.2 x$

$\frac{d x}{10 - 0.2 x} = d t$

Intergrating both sides

$\frac{l o g (10 - 0.2 x)}{(- 0.2)} = t + C$

x(0) = 1

$\Rightarrow C = \frac{l o g 9.8}{(- 0.2)}$

$\frac{l o g (10 - 0.2 x)}{(- 0.2)} - \frac{l o g 9.8}{(- 0.2)} = t$

$l o g \frac{(10 - 0.2 x)}{9.8} = - 0.2 t$

x(t) = 50 - $49 e^{- 0.2 t}$

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Consider the differential equation dxdt=10−0.2x with initial condition x(0) = 1, The response x(t) for t > 0 is

The correct option is C 50−49e−0.2tdxdt=10−0.2x dx10−0.2x=dt Intergrating both sides log(10−0.2x)(−0.2)=t+C x(0) = 1 ⇒C=log9.8(−0.2) log(10−0.2x)(−0.2)−log9.8(−0.2)=t log(10−0.2x)9.8=−0.2t x(t) = 50 -49e−0.2t

Consider the differential equation $\frac{d x}{d t} = 10 - 0.2 x$ with initial condition x(0) = 1, The response x(t) for t > 0 is