Question

how to transform this dy/dt-y=t+1 to it"s general sol?

Answer 1

this is linear eq

Answer 2

http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 3

One thing I would so is just guess that it is a polynomial. \[
y = At+B \\
y' = A \\
y'-y = A - (At+B) = -At + A - B \\
y'-y = t+1 = -At + A-B \implies (1)t + (1) = (-A)t + (A-B)
\]Then try to solve for our coefficients: \[
1 = -A \\
1 = A-B
\]

Answer 4

Next, we want to find the solution in the homogeneous case: \[
y' - y = 0
\]For first order differential equations, we will be in the form: \[
y' + f(t) y = 0
\]And the homogeneous solution is just: \[
e^{\int f(t)dt}
\]

Answer 5

So add up the homogeneous and non-homogeneous parts, to get: \[
Ce^{t}-t-2
\]

Answer 6

I meant to write \[
e^{-\int f(t)dt}
\]

Answer 7

@wio why -ve there at the integrating facto?should be positive right?

Answer 8

Oh, yeah, it should be positive, I was mixing up: \[
y' +f(t) y= 0
\]with \[
y' = f(t)y
\]

Answer 9

\[ \large
\mu (t) = e^{\int-dt} = e^{-t} \]\[ \large
\begin{array}{rc'}
e^{-t}y' - e^{-t}y &=& e^{-t}(t+1) \\
(e^{-t}y)' &=& te^{-t} + e^{-t} \\
e^{-1}y &=& (-te^{-t}-e^{-t}) + (-e^{-t}) \\
y &=& -t - 2
\end{array}
\]

Answer 10

Weird............

Answer 11

dy/dt = y+t+1

y+t+1 = v
dv/dt = dy/dt+ 1
=> dy/dt = dv/dt -1

replacing, we get
dv/dt -1 = v
dv/dt = v+1

dv/(v+1) = dt

integrate to get
ln(v+1) = t +c
ln(y+t+1) = t+c
y+t+1 = ke^t
y = ke^t -t -1

Answer 12

@FoolAroundMath thanks for the help, but one thing doesn't add up: \[
y = ke^t-t-1 \\
y' = ke^t-1 \\
y' - y = (ke^t-1) - (ke^t-t-1) = ke^t-1 - ke^t + t + 1 = t = t \neq t+1
\]

Answer 13

yeah my bad... made a mistake upon substitution: reposting
dy/dt = y+t+1

y+t+1 = v
dv/dt = dy/dt+ 1
=> dy/dt = dv/dt -1

replacing, we get
dv/dt -1 = v
dv/dt = v+1

dv/(v+1) = dt

integrate to get
ln(v+1) = t +c
ln(y+t+2) = t+c
y+t+2 = ke^t
y = ke^t -t -2