# De 17 y'-2y=e^{2t} y(0)=2

• MHB
• karush
In summary: C$is the constant of integration. We solved for$C$earlier, which gave us the decimal approximation.In summary, we used integration by parts to solve the differential equation$ty' + 2y = \sin{t}$and found the particular solution$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{\pi^2-4}{4t^2}$. We then used the initial condition$y\left(\dfrac{\pi}{2}\right) = 1$to solve for the constant of integration$C$, which gave us the exact solution$y = -\dfrac
karush
Gold Member
MHB
#18
$ty'+2y=\sin t \quad y(\pi/2)=1 \quad t\ge 0$
$y'+\dfrac{2}{t}y=\dfrac{\sin t}{t}$
so
$u(x) = e^{\int 2/t dt}= e^{t^2/4}$

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karush said:
#18
$ty'+2y=\sin t \quad y(\pi/2)=1 \quad t\ge 0$
$y'+\dfrac{2}{t}y=\dfrac{\sin t}{t}$
so
$u(x) = e^{\int 2/t dt}= e^{t^2/4}$
$u(x) = e^{\int (2/t) dt}= e^{2\ln t} = \left(e^{\ln t}\right)^2 = t^2$.

Opalg said:
$u(x) = e^{\int (2/t) dt}= e^{2\ln t} = \left(e^{\ln t}\right)^2 = t^2$.

$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ?

karush said:
$t^2 y'+\dfrac{2t^2}{t}y=\dfrac{t^2\sin t}{t}$
$t^2 y'+2ty=t\sin t$
$(yt^2)'=t\sin t$

proceed ?
Yes, keep going! (Yes) (Integrate both sides with respect to $t$.)

Opalg said:
Yes, keep going! (Yes) (Integrate both sides with respect to $t$.)
$(yt^2)'=t\sin t$
IBP gives
$yt^2=-t\cos (t)+\sin (t)+C$
isolate y and reduce
$y=-\dfrac{\cos (t)}{t}+\dfrac{\sin (t)}{t^2}+\dfrac{c}{t^2}$ if ok next $y(\pi/2)=1$

isn't $t^2$ going to be ?

Last edited:
$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y\left(\dfrac{\pi}{2}\right) = 1$

$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$

solve for $C$

skeeter said:
$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{C}{t^2}$

$y\left(\dfrac{\pi}{2}\right) = 1$

$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$

solve for $C$

$1 = 0 + \dfrac{4}{\pi^2} + \dfrac{4C}{\pi^2}$
$\pi^2=4+4C$
$C=\dfrac{\pi ^2-4}{4}\approx1.4674$wasn't expecting a decimal but..not sure how you would ck this?

what's up with the decimal? keep the exact value for $C$

$y = -\dfrac{\cos{t}}{t} + \dfrac{\sin{t}}{t^2} + \dfrac{\pi^2-4}{4t^2}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \dfrac{4}{\pi^2} + \dfrac{\pi^2-4}{4 \cdot \frac{\pi^2}{4}}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \dfrac{4}{\pi^2} + \dfrac{\pi^2-4}{\pi^2}$

$y\left(\dfrac{\pi}{2}\right) = 0 + \cancel{\dfrac{4}{\pi^2}} + \dfrac{\pi^2}{\pi^2} - \cancel{\dfrac{4}{\pi^2}}$

$y\left(\dfrac{\pi}{2}\right) = 1$

$+ \dfrac{\pi^2-4}{4t^2}$

Where did this come from?

karush said:
$+ \dfrac{\pi^2-4}{4t^2}$

Where did this come from?

it’s $\dfrac{C}{t^2}$

## 1. What is the meaning of the equation "De 17y'-2y=e^{2t} y(0)=2"?

The equation represents a differential equation, where the derivative of y (represented by y') is multiplied by a constant (17) and then subtracted from 2 times y. The equation also includes an initial condition, where the value of y at t=0 is equal to 2.

## 2. What is the purpose of using a differential equation in this form?

Differential equations are used to model real-world phenomena that involve rates of change. In this case, the equation represents a relationship between the rate of change of y and the function e^{2t}. Solving this equation can help us understand how y changes over time.

## 3. How can we solve this differential equation?

There are various methods for solving differential equations, such as separation of variables, integrating factors, and substitution. In this case, we can use the integrating factor method to solve for y.

## 4. What is the significance of the initial condition y(0)=2?

The initial condition represents the starting point for the solution of the differential equation. In this case, it tells us that when t=0, the value of y is equal to 2. This helps us determine the specific solution to the equation.

## 5. Can this equation be used to model a real-world situation?

Yes, this equation can be used to model many real-world situations, such as population growth, radioactive decay, and chemical reactions. The specific interpretation of the equation would depend on the context in which it is used.

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