General solution of the differential equation y ′ = y x + x e x ?

Name: General solution of the differential equation y ′ = y x + x e x ?
Uploaded: 2022-02-23T17:22:57+00:00
Description: General solution of the differential equation y ′ = y x + x e x ?

We seek a solution to the First Order ODE: y ′ = y x + x e x We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form; d y d x + P ( x ) y = Q ( x ) So rewrite the equations in standard form as: d y d x - y x = x e x ... . . [ 1 ] Then the integrating factor is given by; I = e ∫ P ( x ) d x = exp ( ∫ - 1 x d x ) = exp ( - ln x ) = 1 x And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential; 1 x d y d x - y x 2 = e x ∴ d d x ( y x ) = e x Which we can directly integrate to get: y x = ∫ e x + C ∴ y x = e x + C ∴ y = x e x + C x

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First order differential equations

rustenig

Answered question

2022-09-06

General solution of the differential equation

y' = \frac{y}{x} + x e^{x}

Answer & Explanation

Raphael Singleton

Beginner2022-09-07Added 19 answers

We seek a solution to the First Order ODE:

y' = \frac{y}{x} + x e^{x}

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

\frac{d y}{d x} + P (x) y = Q (x)

So rewrite the equations in standard form as:

\frac{d y}{d x} - \frac{y}{x} = x e^{x} ... . . [1]

Then the integrating factor is given by;

I = e^{\int P (x) d x}

= \exp (\int - \frac{1}{x} d x)

= \exp (- \ln x)

= \frac{1}{x}

And if we multiply the DE [1] by this Integrating Factor, I, we will have a perfect product differential;

\frac{1}{x} \frac{d y}{d x} - \frac{y}{x^{2}} = e^{x}

∴ \frac{d}{d x} (\frac{y}{x}) = e^{x}

Which we can directly integrate to get:

\frac{y}{x} = \int e^{x} + C

∴ \frac{y}{x} = e^{x} + C

∴ y = x e^{x} + C x

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