- #1
karush
Gold Member
MHB
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well each one is a little different so,,,
$$\dfrac{dy}{dt}=\dfrac{ty(4-y)}{3},\qquad y(0) =y_0$$
not sure if this is what they meant on the given expression
ok yes the 2,2 section exercises are all on separable equationsJoppy said:Could you be clear on what you're struggling with? For instance a) requires first the solution of the DE. Can you identify the kind of DE that it is and suggest a solution approach? (e.g., separable)
$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$Alan said:You just need to integrate $$dy/(y(4-y))=tdt/3$$ by using the partial fractions $1/(y(4-y))=A/y+B/(4-y)$ where you find A and B by multiplying first by $y$ and then plugging y=0 to find A, as for B just try to multiply by 4-y and again plug $y=4$ to find the coefficients A and B. Thus you immediate integrals to solve.
You are off by a minus sign.karush said:$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!
$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$ thus if so then $$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy =\dfrac{1}{3} \int t\, dt $$ |
karush said:
$\displaystyle \dfrac{1}{y(4¬y)} = \dfrac{1}{4y} - \dfrac{1}{4(y - 4)}$
thus
if so then
$$\dfrac{1}{4} \displaystyle\int \dfrac{1}{y} \, dy- \dfrac{1}{4} \int\dfrac{1}{y-4}\, dy
=\dfrac{1}{3} \int t\, dt $$
Which ugly box are you talking about?karush said:$\dfrac{1}{(4 - y) y} = \dfrac{ 1} {4 y} - \dfrac{1}{4 (y - 4)}$
like this?
do we really need this ugly brown box at the bottom!
on a tablet the brown footer is a huge retangle with ears on the sidesAlan said:Which ugly box are you talking about?
OK, I am using a Desktop and this ugly box isn't shown there.karush said:on a tablet the brown footer is a huge retangle with ears on the sides
terrifying to stare at
y_0 is a constant and t is a variable, so this doesn't make so much sense.karush said:$$\displaystyle\int \dfrac{1}{y} \, dy- \int\dfrac{1}{y-4}\, dy
=\dfrac{4}{3} \int t\, dt$$
$$\ln |y|-\ln |y-4|=\frac{2t^2}{3}+C$$
ok assume find C from RHS and $y(0) =y_0$
does $y_0=t$??
A first order initial value problem is a type of differential equation that involves finding a function that satisfies both a given differential equation and a specified initial condition. The initial condition typically involves a value for the function at a specific point.
A first order IVP is typically solved using analytical or numerical methods. Analytical methods involve finding an exact solution to the differential equation, while numerical methods involve approximating the solution using a computer algorithm.
An explicit solution to a first order IVP is one where the dependent variable (usually denoted by y) is isolated on one side of the equation. An implicit solution, on the other hand, is one where the dependent variable appears on both sides of the equation.
A particular solution to a first order IVP is a specific solution that satisfies both the differential equation and the initial condition. This solution is unique for a given initial condition, while the general solution may have multiple solutions.
First order IVPs are used to model a wide range of real-world phenomena, including population growth, chemical reactions, and electrical circuits. They are also used in physics and engineering to describe the behavior of systems over time.