Solve differential equation (2y−ex)dx+xdy=0, x&gt;0

(2y−ex)dx+xdy=0 2y−ex+xdy/dx=0 (2y)/x−ex/x+dy/dx=0 dy/dx+(2/x)y=ex/x Above differential equation is form of dy/dx+Py=Q Integrating factor will be I.F.=e∫Pdx =e∫2/xdx =e2ln⁡x =eln⁡x2 =x2 Solution will be given as I.F∗y=∫I.F∗Qdx x2y=∫x2∗e2/xdx x2y=∫xexdx x2y=[x∫exdx−∫[d/dx(x)∫exdx]]dx x2y=[xex−ex]+c y=1/x2(xex−ex+c)

Question: "Solve differential equation (2y−ex)dx+xdy=0, x>0"

College
Calculus and Analysis
Differential Equations
First order differential equations

melodykap

Answered question

2020-11-10

Solve differential equation

(2 y - e^{x}) d x + x d y = 0

, x>0

Answer & Explanation

Nathanael Webber

Skilled2020-11-11Added 117 answers

$(2 y - e^{x}) d x + x d y = 0$
$2 y - e^{x} + x d y / d x = 0$
$(2 y) / x - e^{x} / x + d y / d x = 0$
$d y / d x + (2 / x) y = e^{x} / x$
Above differential equation is form of $d y / d x + P y = Q$
Integrating factor will be
$I . F . = e^{\int P d x}$
$= e^{\int 2 / x d x}$
$= e^{2 \ln x}$ $= e^{\ln x^{2}}$
$= x^{2}$
Solution will be given as
$I . F * y = \int I . F * Q d x$
$x^{2} y = \int x^{2} * e^{2} / x d x$
$x^{2} y = \int x e^{x} d x$
$x^{2} y = [x \int e^{x} d x - \int [d / d x (x) \int e^{x} d x]] d x$
$x^{2} y = [x e^{x} - e^{x}] + c$
$y = 1 / x^{2} (x e^{x} - e^{x} + c)$

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