solve y' = (10yx - y) / (5x^2 - x) with y(1) = 8

[Guest]

Find the solution to the differential equation y' = (10yx - y) / (5x^2 - x)
that satisfies the initial condition y(1) = 8.

I tried separating the x's and y's and got:

. . .y' = y(10x - 1) / (5x^2 - x)

Then I tried separating the equations and got:

. . .dy/y = [(10x - 1)dx] / (5x^2 - x)

Then I integrated both sides and got:

. . .ln(y) = log(x) + log(5x - 1) + c

Then I solved for y and got:

. . .y = e^(log(x) + log(5x - 1) + c)

. . .y = e^(log(x) + log(5x - 1)) * e^(c)

. . .y = c e^(log(x) + log(5x - 1)

Then I solved for x = 1 and y = 8 and got:

. . .8 = c * 1.826

. . .c = 4.381

So the final soution would be:

. . .y = 4.381e^(log(x)+log(5x-1))

I'm not sure if this is correct. I figured the constant c wouldn't be a decimal like this.

 

[stapel]

You can check by differentiating back, but probably it would be good to at least check the initial condition.

. . . . .y = 4.381 e^{(ln(x) + ln(5x - 1))}

. . . . .y(1) = 4.381 e^{(ln(1) + ln(5 - 1))}

. . . . . . . . .= 4.381 e^{(0 + ln(4))}

. . . . . . . . .= 4.381 e^ln(4)

. . . . . . . . .= (4.381)(4)

...which does not equal 8.

You have:

. . . . .y' = (10xy - y) / (5x² - x)

Then:

. . . . .dy/dx = [y(10x - 1)] / [5x² - x]

. . . . .(1/y) dy = [(10x - 1) / (5x² - x)] dx

. . . . .ln(y) = int [1/x + 5/(5x - 1)] dx

. . . . .ln(y) = ln(x) + ln(5x - 1) + C

. . . . .y = e^{ln(x) + ln(5x - 1) + C}

. . . . . . . . .= [e^ln(x)] [e^{ln(5x - 1)}] [e^C]

. . . . . . . . .= [x] [5x - 1] [D]

. . . . . . . . .= D(5x² - x)

The initial condition gives:

. . . . .y(1) = D(5 - 1) = 4D = 8

. . . . .D = 2

Then:

. . . . .y = 2(5x² - x)

Checking:

. . . . .y(1) = 2(5 - 1) = 8

So that works. Also:

. . . . .y' = 2(10x - 1)

. . . . . . . . .= 20x - 2

Plugging the "y" above into the original equation, we get:

. . . . .y' = [10x(2(5x² - x)) - (2(5x² - x))] / [5x² - x]

. . . . . . . . .= [20x(5x² - x) - 2(5x² - x)] / [5x² - x]

. . . . . . . . .= [20x - 2] / 1

. . . . . . . . .= 20x - 2

So the two derivatives simplify to the same expression.

Hope that helps a bit.

Eliz.

 

[galactus]

\(\displaystyle \L\\y'=\frac{10xy-y}{5x^{2}-x}\)

Factor:

\(\displaystyle \L\\y'=\frac{y(10x-1)}{x(5x-1)}\)

Separate variables:

\(\displaystyle \L\\\frac{1}{y}y'=\frac{10x-1}{x(5x-1)}\)

Expand:

\(\displaystyle \L\\\frac{1}{y}y'=\frac{5}{5x-1}+\frac{1}{x}\)

Integrate:

\(\displaystyle \L\\ln(y)=ln(x)+ln(5x-1)+C\)

\(\displaystyle \L\\ln(y)=ln(x(5x-1))+C\)

e to both sides to solve for y:

\(\displaystyle \L\\y=e^{c}x(5x-1)\)

Use IC: y(1)=8

\(\displaystyle \L\\y=e^{c}(1)(5(1)-1)=8\)

\(\displaystyle c=ln(2)\)

This gives us:

\(\displaystyle \H\\y=2x(5x-1)=10x^{2}-2x\)

Uh-oh, Stapel beat me. We get the same thing, so I would suspect we're correct.

 

[Guest]

That helps alot, thank you both!

 

[soroban]

Hello, almostacasper!

Find the solution to the differential equation: \(\displaystyle \,y' \:= \:\frac{10yx\,-\,y}{5x^2\,-\,x}\)
that satisfies the initial condition \(\displaystyle y(1) \,= \,8\)

Separate variables: \(\displaystyle \L\:\frac{dy}{y}\:=\:\frac{10x\,-\,1}{5x^2\,-\,x}\,dx\)

Integrate: \(\displaystyle \L\:\int\frac{dy}{y}\;=\;\int\frac{10x\,-\,1}{5x^2\,-\,x}\,dx\)

. . Didn't anyone notice that the second integral is: \(\displaystyle \L\int\frac{du}{u}\) ?


So we have: \(\displaystyle \L\:\ln|y| \;= \;\ln|5x^2\,-\,x|\,+\,c \;= \;\ln|5x^2\,-\,x|\,+\,\ln C \;=\;\ln|C(5x^2\,-\,x)|\)

. . Then: \(\displaystyle \L\:y \;=\;C(5x^2\,-\,x)\)


Since \(\displaystyle y(1)\,=\,8\), we have: \(\displaystyle \:C(5\cdot1^2\,-\,1)\:=\:8\;\;\Rightarrow\;\;C\,=\,2\)


Therefore: \(\displaystyle \L\:y\;=\;2(5x^2\,-\,x)\)