Question

Use separation of variables to find the general solution of the differential equation.
y' − xy = 0

Answer 1

so, "separate the variables" :p !!!

Answer 2

Well yeah haha, i got up to y'/y=x, and then i think the next step is (dy/dx)y=x which is then (dy)y=x(dx) but i'm completely lost after that

Answer 3

Now you can just treat those both as integrals, add the little "S" and don't forget the constant of integration

Answer 4

So then eln|y|=e^x2/2 +e^c which is just y=e^x2/2 +e^C I don't know it seems off to me..

Answer 5

\(\large\color{black}{ \displaystyle \frac{dy}{dx}=xy }\)

\(\large\color{black}{ \displaystyle \frac{1}{y}\frac{dy}{dx}=x }\)

\(\large\color{black}{ \displaystyle \color{red}{\int}\frac{1}{y}\frac{dy}{dx}~\color{red}{dx}~=\color{red}{\int}x~\color{red}{dx} }\)

\(\large\color{black}{ \displaystyle \color{red}{\int}\frac{1}{y}\frac{dy}{\cancel{dx}}~\color{red}{\cancel{dx}}~=\color{red}{\int}x~\color{red}{dx} }\)

\(\large\color{black}{ \displaystyle \int \frac{1}{y}~dy=\int x~dx }\)

\(\large\color{black}{ \displaystyle \ln(y)+C=\frac{1}{2}x^2+C }\)

\(\large\color{black}{ \displaystyle \ln(y)=\frac{1}{2}x^2+C }\)

\(\large\color{black}{ \displaystyle e^{\ln(y)}=e^{^{\LARGE{\frac{1}{2}x^2+C}}} }\)

\(\large\color{black}{ \displaystyle y=e^{^{\LARGE{\frac{1}{2}x^2+C}}} }\)

\(\large\color{black}{ \displaystyle y=e^{^{\LARGE{\frac{1}{2}x^2}}}e^C }\)

\(\large\color{black}{ \displaystyle y=Ce^{^{\LARGE{\frac{1}{2}x^2}}} }\)

Answer 6

oh my gosh THANK YOU!

Answer 7

yw.....