Find the solution of the given initial value problem.

${y}^{\prime}-2y={e}^{2t},y(0)=2$

${y}^{\prime}-2y={e}^{2t},y(0)=2$

anudoneddbv
2022-07-30
Answered

Find the solution of the given initial value problem.

${y}^{\prime}-2y={e}^{2t},y(0)=2$

${y}^{\prime}-2y={e}^{2t},y(0)=2$

You can still ask an expert for help

Killaninl2

Answered 2022-07-31
Author has **20** answers

Given equation is ${y}^{1}-2y={e}^{2t}$

This is a linear equation of first order in y

where P(t)=-2 and $Q(t)={e}^{2t}$

$\int P(t)dt=-2\int 1\ast dt=-2t$

Integrating Factor $={e}^{\int P(t)dt}={e}^{-2t}$

Multiplying (1) with ${e}^{-2t}$, we get

${e}^{2t}{y}^{1}-2{e}^{-2t}y=1\to (2)$

(2) can be written as

$\frac{d}{dt}[y\ast {e}^{-2t}]dt=\int 1\ast dt$

$\Rightarrow y\ast {e}^{-2t}=t+C$

$\Rightarrow y(t)=\frac{t}{{e}^{-2t}}+\frac{C}{{e}^{-2t}}$

we know that y(0)=2

so $\frac{0}{{e}^{-0}}+\frac{C}{{e}^{-0}}=2\Rightarrow C=2$

Therefore the general solution of (1) is $y=\frac{t+2}{{e}^{-2t}}$

This is a linear equation of first order in y

where P(t)=-2 and $Q(t)={e}^{2t}$

$\int P(t)dt=-2\int 1\ast dt=-2t$

Integrating Factor $={e}^{\int P(t)dt}={e}^{-2t}$

Multiplying (1) with ${e}^{-2t}$, we get

${e}^{2t}{y}^{1}-2{e}^{-2t}y=1\to (2)$

(2) can be written as

$\frac{d}{dt}[y\ast {e}^{-2t}]dt=\int 1\ast dt$

$\Rightarrow y\ast {e}^{-2t}=t+C$

$\Rightarrow y(t)=\frac{t}{{e}^{-2t}}+\frac{C}{{e}^{-2t}}$

we know that y(0)=2

so $\frac{0}{{e}^{-0}}+\frac{C}{{e}^{-0}}=2\Rightarrow C=2$

Therefore the general solution of (1) is $y=\frac{t+2}{{e}^{-2t}}$

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As far as I know this sort of dynamical system may have no closed-form solution, but even if it has one, the $=$ replaced in it by $\le $ would include more sequences than the above recurrence allows.