Question

Solve the differential equation.

Answer 1

\[xy'+y=37\sqrt{x}\]

Answer 2

\[y ' + y/x = 37/\sqrt{x} = > y ' + y(1/x) = 37/\sqrt{x}\]
\[e \int\limits_{}^{}1^{(1/x)dx} = e^{lnx}=x\]\[( y ' + (\frac{ 1 }{ x })y) * x = \frac{ 37 }{ \sqrt{37} }*x\]\[x y ' + y = 37 \sqrt{x}\]

What I have so far

Answer 3

yes, which is what you started with, but now you know
the left-hand side is an exact differential

Answer 4

the left side is (xy)' = x dy/dx + y dx/dx = x dy/dx + y or x y ' + y

Answer 5

dx/dx means differentiating in terms of y?

Answer 6

just like how dy/dx is differentiating with respect to x.

Answer 7

I was showing
\[ \frac{d}{dx} x = 1 \]
the derivative of x with respect to x is 1
which is the first problem you learn in calculus
but it's helpful (to me) if I write it as dx/dx to remember it is 1
and though y' is easy to type, it's helpful to remember it stands for dy/dx

Answer 8

okay,\[x*\frac{ dy}{ dx }+y*\frac{ dx }{ dx }=37x^{1/2}\]\[\frac{ d(u*v) }{ dx }= u * \frac{ dv }{ dx }+v*\frac{ du }{ dx }\]\[\frac{ d(x*y) }{ dx }=37x^{1/2} \]

How would I proceed from there?

Answer 9

write it as
\[ d(xy) = 37 x^\frac{1}{2} \ dx \\
\int d(xy) = 37 \int x^\frac{1}{2} \ dx \\
xy = 37 \int x^\frac{1}{2} \ dx
\]

Answer 10

How does \[\int\limits_{}^{}d(xy) = xy\]?

Answer 11

by definition
\[ \int d x = x \]
if x is "complicated" e.g. xy , it still works

Answer 12

So, \[\int\limits_{}^{}d(xy)\] basically means \[\int\limits_{}^{}dx *\int\limits_{}^{}dy\] = xy
?

Answer 13

no. it means
\[ \int x\ dy + y\ dx \]

Answer 14

which may not be obvious
I think of it as integral of the derivative are inverse operations. and we "drop" both of them
\[ \int d(xy) = xy\]

Answer 15

Okay, got it. I'll just think of it like that then.

Answer 16

\[xy = \frac{ 74 }{ 3 }x^{3/2} + C\]\[xy = \frac{ 74x \sqrt{x} }{ 3 } + C = > y = \frac{ 74 \sqrt{x} }{ 3 }+\frac{ C }{ x }\]

Answer 17

Thanks for the help again, ~phi

Answer 18

yes, looks good