Question

first order lnear diff eq

Answer 1

my integrating factor would be x^-2

Answer 2

@johnweldon1993

Answer 3

What kind of linear equation is this?
\[y'=P(x)y+Q(x)\quad\text{or}\quad y'+P(x)y=Q(x)~~?\]

Answer 4

I only ask because in your drawing you have two equal signs...

Answer 5

dy/dx + P(x)y = Q(x)

Answer 6

no i mean.. dy/dx- 2y/x = x^2 e^x im sorry

Answer 7

\[\frac{dy}{dx}-\frac{2}{x}y=x^2e^x\]
Integrating factor:
\[\ln \mu(x)=\int -\frac{2}{x}\,dx~~\implies~~\mu(x)=x^{-2}\]
Distributing the IF, you have
\[\begin{align*}
\frac{1}{x^2}\frac{dy}{dx}-\frac{2}{x^3}y&=e^x\\\\
\frac{d}{dx}\left[\frac{1}{x^2}y\right]&=e^x\\\\
\frac{1}{x^2}y&=\int e^x\,dx\\\\
\frac{1}{x^2}y&=e^x+C\\\\
y&=x^2e^x+Cx^2
\end{align*}\]
Knowing that \(y=0\) when \(x=1\), yo can then find a value for \(C\).

Answer 8

c= 2.72

Answer 9

c= -2.72

Answer 10

Right, it's the negative one. Typically, you're better off with the exact form, \(C=-e\).

Answer 11

yezzz thank you so muccch