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Homework Statement
##\int \frac{d}{dx} (x^2y)##
?
Explain please and give me some hints.
You are supposed to show an attempt at solving. Also you haven't told us several things. What variable is the integration with respect to? Is ##y## a function of ##x## or an independent variable?Homework Statement
##\int \frac{d}{dx} (x^2y)##
?
Explain please and give me some hints.
Homework Equations
The Attempt at a Solution
You are supposed to show an attempt at solving. Also you haven't told us several things. What variable is the integration with respect to? Is ##y## a function of ##x## or an independent variable?
You are supposed to show an attempt at solving. Also you haven't told us several things. What variable is the integration with respect to? Is ##y## a function of ##x## or an independent variable?
Is ##\int \frac{d}{dx} (x^2y) = x^2y \ ?##
So show us your work so far
Solve
##\frac{dy}{dx} - 3y = 0.##
Solution
This linear equation can be solved by separation of variables. Alternatively, since the equation is already in the standard form ##[\frac{dy}{dx} + P(x) \ y = f(x)]##, we see that ##P(x) = - 3## and so the integrating factor is
##e^{\int P(x) \ dx} = e^{\int (-3) \ dx} = e^{-3x}.##
We multiply the equation by this factor and recognize that
##e^{-3x} \frac{dy}{dx} - 3 e^{-3x} y = 0## is the same as ##\frac{d}{dx}[e^{-3x}y] = 0.##
Integrating both sides of the last equation gives ##e^{-3x}y = c.## Solving for ##y## gives us the explicit solution ##y = ce^{3x}, -∞ < x < ∞.##
This is the same type of problem you were stuck on yesterday.basty said:We multiply the equation by this factor and recognize that
##e^{-3x} \frac{dy}{dx} - 3 e^{-3x} y = 0## is the same as ##\frac{d}{dx}[e^{-3x}y] = 0.##
Solve
##\frac{dy}{dx} - 3y = 6.##
Solution
The associated homogeneous equation for this DE was solved in above example. Again the equation is already in the standard form, and the integration factor is still ##e^{\int (-3) dx} = e^{-3x}.## This time multiplying the given equation by this factor gives
##e^{-3x}\frac{dy}{dx} - 3 e^{-3x}y = 6e^{-3x},## which is the same as ##\frac{d}{dx}[e^{-3x}y] = 6e^{-3x}##
Integrating both sides of the last equation gives ##e^{-3x}y = -2e^{-3x} + c## or ##y = -2 + ce^{3x}, - ∞ < x < ∞.##
Again it shows thatSolve
##x \frac{dy}{dx} - 4y = x^6e^x.##
Solution
Dividing by ##x,## we get the standard form
##\frac{dy}{dx} - \frac{4}{x}y = x^5e^x.##
From this form we identify ##P(x) = -\frac{4}{x}## and ##f(x) = x^5e^x## and further observe that ##P## and ##f## are continuous on ##(0, ∞).## Hence the integrating factor is
##e^{-4 \int \frac{dx}{x}} = e^{-4 \ln x} = e^{\ln x^{-4}} = x^{-4}##
Now we multiply ##\frac{dy}{dx} - \frac{4}{x}y = x^5e^x## by ##x^{-4}## and rewrite
##x^{-4}\frac{dy}{dx} - 4x^{-5}y = xe^x## as ##\frac{d}{dx}[x^{-4}y] = xe^x##
It follows from integration by parts that the general solution defined on the interval ##(0, ∞)## is
##x^{-4}y = xe^x - e^x + c## or ##y = x^5e^x - x^4e^x + cx^4.##
If d/dx(?) = 0, what does ? have to be?
Actually, from the text, it's clear the author means to integrate with respect to ##x##. That is, you haveI rewrite the example from my differential equations book into this thread, below.
When he says "integrating both sides of the last equation..." mean ##\int \frac{d}{dx}[e^{-3x}y] = \int 0.##
##\int \frac{d}{dx}(whatever) = whatever##
Similar thing in another examples he provides in the book.
These results are from beginning calculus. It seems a little strange that you're taking differential equations if you're not familiar with these facts already.I don't understand. Please explain more.If d/dx(?) = 0, what does ? have to be?
If d/dx(?) = 0, what does ? have to be?
I don't understand. Please explain more.
I agree with vela. You have started a number of threads with questions about differential equations, but have run into roadblocks in concepts from algebra, trig, and basic calculus. You will not be successful in studying differential equations if you don't understand the basic concepts that you should already be competent in.These results are from beginning calculus. It seems a little strange that you're taking differential equations if you're not familiar with these facts already.