Question

Eliza Beth13 · Accepted Answer

To solve the given first-order differential equation (DE), we'll use the method of integrating factors. The DE is as follows:

t y^{'} + (t + 1) y = 2 t e^{- t} with y (1) = a

Rewrite the DE in standard form
To simplify the equation, we divide both sides by

t

:

y^{'} + (1 + \frac{1}{t}) y = 2 e^{- t}

Identify the integrating factor
The integrating factor for this DE is given by the exponential of the integral of the coefficient of

y

, which is

(1 + \frac{1}{t})

. Let's calculate it:

\int (1 + \frac{1}{t}) d t = \int \frac{t + 1}{t} d t = \int (1 + \frac{1}{t}) d t = t + \ln | t | + C

So, the integrating factor is

e^{t + \ln | t |} = t e^{t}

.
Multiply the DE by the integrating factor
Multiply both sides of the DE by

t e^{t}

:

t e^{t} y^{'} + t e^{t} (1 + \frac{1}{t}) y = 2 t e^{t} e^{- t}

Simplifying, we have:

t e^{t} y^{'} + (t + 1) y e^{t} = 2 t

Apply the product rule on the left side
To apply the product rule, let's define

u (t) = y e^{t}

. Now, we can differentiate

u (t)

:

\frac{d}{d t} (u (t)) = \frac{d}{d t} (y e^{t})

Using the product rule, we get:

u^{'} (t) = y^{'} e^{t} + y e^{t}

Rewrite the DE using

u (t)

Substituting the expressions from Step 4 into the DE, we have:

t e^{t} (u^{'} (t)) + (t + 1) u (t) = 2 t

Solve the resulting linear DE
The DE obtained in Step 5 is linear, so we can solve it using standard methods. First, let's rearrange the equation:

t e^{t} u^{'} (t) + (t + 1) u (t) = 2 t

Integrate both sides
Integrating both sides of the equation, we obtain:

\int t e^{t} u^{'} (t) d t + \int (t + 1) u (t) d t = \int 2 t d t

Evaluate the integrals
Let's calculate the integrals one by one:

\int t e^{t} u^{'} (t) d t = \int u (t) d u (using substitution)

\int (t + 1) u (t) d t = \int (t + 1) (y e^{t}) d t

\int 2 t d t = t^{2}

Simplify the equation
Substituting the results back into the equation, we have:

u (t) + \int (t + 1) (y e^{t}) d t = t^{2}

Evaluate the remaining integral
Let's calculate the remaining integral:

\int (t + 1) (y e^{t}) d t

Expanding the expression and using integration by parts, we get:

\int (t + 1) (y e^{t}) d t = \int (t y e^{t} + y e^{t}) d t

Using integration by parts, with

v = e^{t}

and

d u = (t + 1) y e^{t} d t

, we find:

\int (t + 1) (y e^{t}) d t = (t y e^{t} - \int e^{t} d t) + C

\int (t + 1) (y e^{t}) d t = t y e^{t} - e^{t} + C

Substitute back into the equation
Substituting the result of the integral into the equation

u (t) + (t y e^{t} - e^{t} + C) = t^{2}

Solve for

u (t)

Rearranging the equation, we get:

u (t) = t^{2} - t y e^{t} + e^{t} - C

Solve for

y (t)

Recall that we defined

u (t) = y e^{t}

. Substituting this back into the equation, we have:

y e^{t} = t^{2} - t y e^{t} + e^{t} - C

Simplifying, we get:

y e^{t} + t y e^{t} = t^{2} + e^{t} - C

y e^{t} (1 + t) = t^{2} + e^{t} - C

y = \frac{t^{2} + e^{t} - C}{e^{t} (1 + t)}

Apply the initial condition
To find the value of the constant

C

, we can apply the initial condition

y (1) = a

. Substituting

t = 1

and

y = a

, we get:

a = \frac{1^{2} + e^{1} - C}{e^{1} (1 + 1)}

2 a e = 1 + e - C

C = 1 + e - 2 a e

Final solution
Substituting the value of

C

back into the expression for

y (t)

, we have:

y (t) = \frac{t^{2} + e^{t} - (1 + e - 2 a e)}{e^{t} (1 + t)}

Now, let's determine the range of values of

a

for which

{lim}_{t \to 0^{+}} y = \infty

.
As

t

approaches

0^{+}

, the term

t^{2}

becomes negligible compared to

e^{t}

. Thus, we can ignore the

t^{2}

term. Considering the other terms:

{lim}_{t \to 0^{+}} y (t) = {lim}_{t \to 0^{+}} \frac{e^{t} - (1 + e - 2 a e)}{e^{t} (1 + t)}

As

t

approaches

0^{+}

, the denominator

e^{t} (1 + t)

remains positive and nonzero. Hence, for

{lim}_{t \to 0^{+}} y (t)

to be infinity, the numerator

e^{t} - (1 + e - 2 a e)

must approach infinity.
Since

e^{t}

approaches

1

as

t

approaches

0

, the condition for

{lim}_{t \to 0^{+}} y (t) = \infty

is:

1 - (1 + e - 2 a e) = \infty

Simplifying, we get:

- e + 2 a e = \infty

This condition implies that

- e + 2 a e

should be greater than any finite value, which means that

a

must be greater than

\frac{e}{2}

.
Therefore, the range of values of

a

for which

{lim}_{t \to 0^{+}} y = \infty

is

a > \frac{e}{2}

.

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