Question

Find the general solution for x(dy/dx) - 4y = e^x

Answer 1

Do you know what's the integrating factor?

Answer 2

\[x\frac{dy}{dx} - 4y = e^x\]

Answer 3

The way to solve this is
\[\frac{dy}{dx}+P(x)y=Q(x)\]
Then
\[\Large \mu(x)=e^{\int\limits P(s)ds}\]
once you've got \(\mu (x)\) then just multiply through all the terms

Answer 4

Oh! I thought I had to use a different method, it makes sense now, thank you!

Answer 5

yw :)

Answer 6

find the homogenous parts, then take a guess at the particular parts: Ae^x

Answer 7

didnt see the x in front :)

Answer 8

Heheh ;)

Answer 9

$$xy'-4y=e^x\\y'-\frac4xy=\frac1xe^x$$We want some integration factor \(\mu\) s.t. \(\mu y'-\frac4x\mu y=(\mu y)'=\mu y'+\mu'y\) i.e. \(-\frac4x\mu=\mu'\). $$-\frac4x\mu=\mu'\\-\frac4xdx=\frac1\mu d\mu\\-4\log x=\log\mu\\x^{-4}=\mu$$... s multiplying throughout by \(\mu\) we get $$\frac1{x^4}y'-\frac4{x^5}y=\frac1{x^5}e^x\\\left(\frac1{x^4}y\right)'=\frac1{x^5}e^x\\\frac1{x^4}y=\int\frac1{x^5}e^x\,\mathrm{d}x$$ This last integral is a bit ugly... try using parts:$$\int\underbrace{\frac1{x^5}}_u\,\underbrace{e^x\,\mathrm{d}x}_{\mathrm{d}v}=\frac1{x^5}e^x+\frac14\int\underbrace{\frac1{x^4}}_u\,\underbrace{e^x\mathrm{d}x}_{\mathrm{d}v}=\frac1{x^4}+\frac14\left(\frac1{x^4}e^x+\frac13\int\frac1{x^3}e^x\,\mathrm{d}x\right)=\dots$$