# -2.1.10 solve ty' -y =t^2e^{-1} u(x)

• MHB
• karush
In summary, we can solve the given differential equation by first dividing through by $t$, obtaining a new equation $y'-\frac{y}{t}=te^{-1}$. Then, we can obtain $u(t)$ by taking the exponential of the integral of $-\frac{1}{t}$. Multiplying through by $\frac{1}{t}$, we get a simplified equation $(\frac{y}{t})'=e^{-1}$. Finally, we can integrate and solve for $y$ to get the general solution $y=\frac{t^2}{e}+c_1t$.
karush
Gold Member
MHB
Find the general solution of the given differential equation
$\displaystyle ty^\prime -y =t^2e^{-1}\\$
Divide thru by t
$\displaystyle y^\prime-\frac{y}{t}=te^{-1}\\$
Obtain $u(t)$
$\displaystyle u(t)=\exp\int -\frac{1}{t} \, dt =\frac{1}{t}\\$
Multiply thru with $\displaystyle\frac{1}{t}$
$\displaystyle y^\prime -\frac{1}{t}y =te^{-1}\\$
Simplify:
$(\frac{1}{t}y)'=te^{-1} \\$

ok got stuck here so...

$\displaystyle \color{red}{y=c_1 -te^{-1}}$ (this is wrong)

$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value disappearProblems}$$

Last edited:
karush said:
Find the general solution of the given differential equation
$\displaystyle ty^\prime -y =t^2e^{-1}\\$
Divide thru by t
$\displaystyle y^\prime-\frac{y}{t}=te^{-1}\\$
Obtain $u(t)$
$\displaystyle u(t)=\exp\int -\frac{1}{t} \, dt =\frac{1}{t}\\$
Multiply thru with $\displaystyle\frac{1}{t}$
$\displaystyle y^\prime -\frac{1}{t}y =te^{-1}\\$
Simplify:
$(\frac{1}{t}y)'=te^{-1} \\$

ok got stuck here so...

$\displaystyle \color{red}{y=c_1 -te^{-1}}$

$$\tiny\textsf{Text Book: Elementary Differential Equations and Boundary Value disappearProblems}$$

Hi karush, :)

I think you have done a minor mistake; observe that; $$t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$$.

Hope you can do it from here :)

Ok mucho mahalo

Last edited:
Sudharaka said:
Hi karush, I think you have done a minor mistake; observe that;
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$. Hope you can do it from here
So if
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$
then
$\displaystyle t\left(\frac{y}{t}\right)'=te^{-1}$
or $\displaystyle \left(\frac{y}{t}\right)'=e^{-1}$
then
$\displaystyle \frac{y}{t}=\int e^{-1} dt=\frac{t}{e}+c_1$
multiply thru by t
$\displaystyle y=\frac{t^2}{e}+c_1t$

kinda maybe hopefully raj

Last edited:
karush said:
So if
$\displaystyle t\left(\frac{y}{t}\right)'=y'-\frac{y}{t}$
then
$\displaystyle t\left(\frac{y}{t}\right)'=te^{-1}$
or $\displaystyle \left(\frac{y}{t}\right)'=e^{-1}$
then
$\displaystyle \frac{y}{t}=\int e^{-1} dt=\frac{t}{e}+c_1$
multiply thru by t
$\displaystyle y=\frac{t^2}{e}+c_1t$

kinda maybe hopefully raj

Yes that is correct. Well done. :)

## 1. What does the equation "-2.1.10 solve ty' -y =t^2e^{-1} u(x)" represent?

The equation represents a first-order linear differential equation that can be used to model various physical and biological phenomena.

## 2. What does the term "ty'" in the equation "-2.1.10 solve ty' -y =t^2e^{-1} u(x)" mean?

The term "ty'" represents the derivative of the dependent variable y with respect to the independent variable t.

## 3. How can this equation be solved?

This equation can be solved using various techniques such as separation of variables, integrating factors, and substitution methods.

## 4. What is the significance of the term "e^{-1}" in the equation "-2.1.10 solve ty' -y =t^2e^{-1} u(x)"?

The term "e^{-1}" represents the inverse of the natural logarithm base, which is approximately equal to 0.368. This term is often used in differential equations involving exponential functions.

## 5. Can this equation be used to solve real-world problems?

Yes, this equation can be used to model and solve various real-world problems in fields such as physics, chemistry, biology, and economics.

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