Thread: Initial value problem

Hi, I don't understand how I am meant to solve this:

$\displaystyle x\frac{dy}{dx} + y = x^4, y(1) = 1$

Any help appreciated

Hello, Beard!

$\displaystyle x\frac{dy}{dx} + y \:=\: x^4,\quad y(1) = 1$

Divide by $\displaystyle x\!:\quad \frac{dy}{dx} + \frac{1}{x}\,y \:=\:x^3$

Integrating Factor: .$\displaystyle I \:=\:e^{\int\frac{dx}{x}} \:=\:e^{\ln x} \:=\:x$

Multiply by $\displaystyle x\!:\quad x\,\frac{dy}{dx} + y \:=\:x^4$

We have: .$\displaystyle \frac{d}{dx}(xy) \:=\:x^4$

Integrate: .$\displaystyle xy \:=\:\frac{1}{5}x^5 + C \quad\Rightarrow\quad y \:=\:\frac{1}{5}x^4 + \frac{C}{x}$


From $\displaystyle y(1) = 1\!:\quad 1 \:=\:\tfrac{1}{5}(1^4) + \frac{C}{1} \quad\Rightarrow\quad C \:=\:\frac{4}{5}$


Therefore: .$\displaystyle y \;=\;\frac{1}{5}x^4 + \frac{4}{5x}$

Originally Posted by Soroban

Hello, Beard!


Divide by $\displaystyle x\!:\quad \frac{dy}{dx} + \frac{1}{x}\,y \:=\:x^3$

Integrating Factor: .$\displaystyle I \:=\:e^{\int\frac{dx}{x}} \:=\:e^{\ln x} \:=\:x$

Multiply by $\displaystyle x\!:\quad x\,\frac{dy}{dx} + y \:=\:x^4$

We have: .$\displaystyle \frac{d}{dx}(xy) \:=\:x^4$

Integrate: .$\displaystyle xy \:=\:\frac{1}{5}x^5 + C \quad\Rightarrow\quad y \:=\:\frac{1}{5}x^4 + \frac{C}{x}$


From $\displaystyle y(1) = 1\!:\quad 1 \:=\:\tfrac{1}{5}(1^4) + \frac{C}{1} \quad\Rightarrow\quad C \:=\:\frac{4}{5}$


Therefore: .$\displaystyle y \;=\;\frac{1}{5}x^4 + \frac{4}{5x}$

Thankyou but I am confused by the step "integrating factor" I don't understand how it affects the equation

Originally Posted by Beard

Thankyou but I am confused by the step "integrating factor" I don't understand how it affects the equation

It makes the left-hand side look like the result of a product-rule differentiation. Just in case a picture helps... suppose



is the product rule - straight continuous lines differentiating downwards (integrating up) with respect to x. Then we're trying to fit the left-hand side...

$\displaystyle \frac{dy}{dx} + y\ \frac{1}{x}$

... along the bottom row. As it comes, it won't fit (and satisfy the rule). But if we accomodate the chain rule...



... inside the (legs-uncrossed version of the) product rule...



... we can see how to fix it...

Spoiler:

Hope that helps.

Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

Originally Posted by tom@ballooncalculus

It makes the left-hand side look like the result of a product-rule differentiation. Just in case a picture helps... suppose



is the product rule - straight continuous lines differentiating downwards (integrating up) with respect to x. Then we're trying to fit the left-hand side...

$\displaystyle \frac{dy}{dx} + y\ \frac{1}{x}$

... along the bottom row. As it comes, it won't fit (and satisfy the rule). But if we accomodate the chain rule...



... inside the (legs-uncrossed version of the) product rule...



... we can see how to fix it...

Spoiler:

Hope that helps.

Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

Thanks for the attempt but that has confused me more

Fair enough! Try to focus on the product rule, one way or another... or, maybe not - you could simply practice applying the formula. Whatever works for you...

PS: are you clear, at least, that the equation is 'multiplied through' by the I.F.?

i.e. given...

A + B = C (1)

... it follows that

A * I.F. + B * I.F. = C * I.F. (2)

... and (2) is easier than (1)