How would you solve this (diff eqs) - y'(x) = -3(lambda)y(x)

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Maybe this place will work better

e4

Almost forgot, solving for the case lambda = 1.

>>8820982
shut up and post a lichess game you idiot

[eqn] \implies \mathrm{e}^{3\lambda x}\ y' + 3\lambda \mathrm{e}^{3\lambda x}\ y =\frac{\mathrm{d}}{\mathrm{d}x}\Big(y\mathrm{e}^{3\lambda x}\ \Big) = \Big(\frac{3}{10}\lambda x^2 + \frac{2}{10} x + \lambda \Big) \mathrm{e}^{3 \lambda x} [/eqn]

>>8821006
I only play blindfolded

>>8821012

Can you explain how to do this?

>>8820980
It's time to d-d-d-d-d-d-d-d-d-d-d-d-duel
*unzip horses*

>>8821018
I multiplied both sides by [math] \mathrm{e}^{3 \lambda x} [/math]. Both sides are now easily integrable. The LHS is trivially y times our integrating factor, whereas the RHS is just standard integration by parts.

>>8820980
>I'm supposed to use octave
Then there must be a value for lambda

>>8821031
Sorry, yes, lambda = 1

Sorry, if [math]\lambda = 1[/math] then isn't this just a first order linear equation?

[eqn]
y'(x) = -3y(x) + \frac{3}{10}x^2 + \frac{2}{10}x + 1
[/eqn]

Goes to

[eqn]
y'(x) + p(x)y(x) = g(x)
[/eqn]

where [math]p(x) = 3[/math] and [math]g(x) = \frac{3}{10}x^2 + \frac{2}{10}x + 1[/math]

Easily solvable by normal methods?

>>8821142
They want to know how to do it using a computer program

[math]
y'(x) = -3\lambda y(x) + \frac{3\lambda x^2}{10} + \frac{x}{5} + 1\\
y'(x) + 3\lambda y(x) = \frac{3\lambda x^2}{10} + \frac{x}{5} + 1\\
[/math]

This is a first order linear DE of the form [math] a_0(x)\frac{dy}{dx} + a_1(x)y = g(x)[/math]

[math] a_1(x) = 3\lambda[/math]

The integrating factor is:
[math]
e^{\int{a_1(x) dx}} = e^{\int{3\lambda dx}} = e^{3\lambda x}
[/math]

Multiply the equation by the integrating factor:

[math]

e^{3\lambda x}y'(x) + e^{3\lambda x}3\lambda y(x) = e^{3\lambda x}(\frac{3\lambda x^2}{10} + \frac{x}{5} + 1)
[/math]
now take integral of both sides with respect to x and solve for y.