Given: y′+1xy=cos⁡xSolution:y1+1xy=cos⁡xcomparing with the equationy′+p(x)y=q(x)p(x)=1x,q(x)=cos⁡xM=Integrating factor=e∫p(x)dx=e∫1xdx=eln⁡xm=xgeneral solution is,ym=∫q(x)m dx+cyx=∫cos⁡x⋅x dx+cxy=x∫cos⁡x dx−∫[d(x)dx∫cos⁡x dx]dxxy=xsin⁡x−∫1⋅sin⁡xdxxy=xsin⁡x−∫sin⁡xdxxy=xsin⁡x−(−cos⁡x)+cxy=xsin⁡x+cos⁡x+c

y′+1xy=cos⁡xIntegrating factor P(x)=1xe∫1xdx=eln|x|=xxdydx=xcos⁡x[xy]ddx=xcos⁡xxy=∫xcos⁡xxy=xsin⁡x+cos⁡x+cy=xsin⁡x+cos⁡x+cxy=sin⁡x+cos⁡xx+cx

Given: y′+1xy=cos⁡xSolution:y1+1xy=cos⁡xcomparing with the equationy′+p(x)y=q(x)p(x)=1x,q(x)=cos⁡xM=Integrating factor=e∫p(x)dx=e∫1xdx=eln⁡xm=xgeneral solution is,ym=∫q(x)m dx+cyx=∫cos⁡x⋅x dx+cxy=x∫cos⁡x dx−∫[d(x)dx∫cos⁡x dx]dxxy=xsin⁡x−∫1⋅sin⁡xdxxy=xsin⁡x−∫sin⁡xdxxy=xsin⁡x−(−cos⁡x)+cxy=xsin⁡x+cos⁡x+c

y′+1xy=cos⁡xIntegrating factor P(x)=1xe∫1xdx=eln|x|=xxdydx=xcos⁡x[xy]ddx=xcos⁡xxy=∫xcos⁡xxy=xsin⁡x+cos⁡x+cy=xsin⁡x+cos⁡x+cxy=sin⁡x+cos⁡xx+cx

Step by step solution to "y′+1xy=cos⁡(x)"

Name: y′+1xy=cos⁡(x)
Uploaded: 2022-02-23T17:22:57+00:00
Description: y′+1xy=cos⁡(x)

Gregory Emery

Answered question

2022-01-01

y^{'} + \frac{1}{x} y = \cos (x)

Answer & Explanation

vrangett

Beginner2022-01-02Added 36 answers

Given:

y^{'} + \frac{1}{x} y = \cos x

Solution:

y^{1} + \frac{1}{x} y = \cos x

comparing with the equation

y^{'} + p (x) y = q (x)

p (x) = \frac{1}{x}, q (x) = \cos x

M = Integrating factor = e^{\int p (x) d x}

= e^{\int \frac{1}{x} d x}

= e^{\ln x}

m = x

general solution is,

y_{m} = \int q (x) m d x + c

y_{x} = \int \cos x \cdot x d x + c

x y = x \int \cos x d x - \int [\frac{d (x)}{d x} \int \cos x d x] d x

x y = x \sin x - \int 1 \cdot \sin x d x

x y = x \sin x - \int \sin x d x

x y = x \sin x - (- \cos x) + c

x y = x \sin x + \cos x + c

yotaniwc

Beginner2022-01-03Added 34 answers

y^{'} + \frac{1}{x} y = \cos x

Integrating factor

P (x) = \frac{1}{x}

e^{\int \frac{1}{x} d x} = e^{\ln | x |} = x

x \frac{d y}{d x} = x \cos x

[x y] \frac{d}{d x} = x \cos x

x y = \int x \cos x

x y = x \sin x + \cos x + c

y = \frac{x \sin x + \cos x + c}{x}

y = \sin x + \frac{\cos x}{x} + \frac{c}{x}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

1. A body falls from rest against resistance proportional to the square root of the speed at any instant. If the body attains speed V1 and V2 feet per second, after 1 and 2 seconds in motion, respectively, find an expression for the limiting velocity.
The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True or False
The Laplace transform of $u (t - 2)$ is
(a) $\frac{1}{s} + 2$
(b) $\frac{1}{s} - 2$
(c) $e^{2} \frac{s}{s} (d) \frac{e^{- 2 s}}{s}$ ??
1 degree on celsius scale is equal to
A) $\frac{9}{5}$ degree on fahrenheit scale
B) $\frac{5}{9}$ degree on fahrenheit scale
C) 1 degree on fahrenheit scale
D) 5 degree on fahrenheit scale
The Laplace transform of $t e^{t}$ is A. $\frac{s}{(s + 1)^{2}}$ B. $\frac{1}{(s - 1)^{2}}$ C. $\frac{s}{(s + 1)^{2}}$ D. $\frac{s}{(s - 1)}$
What is the Laplace transform of $t \cos t$ into the s domain?
Find the general solution of the given differential equation:
$y^{″} - 2 y^{'} + y = 0$
The rate at which a body cools is proportional to the difference in
temperature between the body and its surroundings. If a body in air
at 0℃ will cool from 200℃ 𝑡𝑜 100℃ in 40 minutes, how many more
minutes will it take the body to cool from 100℃ 𝑡𝑜 50℃ ?
A body falls from rest against a resistance proportional to the velocity at any instant. If the limiting velocity is 60fps and the body attains half that velocity in 1 second, find the initial velocity.
What's the correct way to go about computing the Inverse Laplace transform of this?
$\frac{- 2 s + 1}{(s^{2} + 2 s + 5)}$
I Completed the square on the bottom but what do you do now?
$\frac{- 2 s + 1}{(s + 1)^{2} + 4}$
How to find inverse Laplace transform of the following function?
$X (s) = \frac{s}{s^{4} + 1}$
I tried to use the definition: $f (t) = L^{- 1} {F (s)} = \frac{1}{2 π i} lim_{T \to \infty} \int_{γ - i T}^{γ + i T} e^{s t} F (s) d s$ or the partial fraction expansion but I have not achieved results.
How do i find the lapalace transorm of this intergral using the convolution theorem? $\int_{0}^{t} e^{- x} \cos x d x$
How can I solve this differential equation? : $x y d x - (x^{2} + 1) d y = 0$
Find the inverse Laplace transform of $\frac{s^{2} - 4 s - 4}{s^{4} + 8 s^{2} + 16}$
inverse laplace transform - with symbolic variables:
$F (s) = \frac{2 s^{2} + (a - 6 b) s + a^{2} - 4 a b}{(s^{2} - a^{2}) (s - 2 b)}$
My steps:
$F (s) = \frac{2 s^{2} + (a - 6 b) s + a^{2} - 4 a b}{(s + a) (s - a) (s - 2 b)}$
$= \frac{A}{s + a} + \frac{B}{s - a} + \frac{C}{s - 2 b} + K$
$K = 0$
$A = F (s) * (s + a)$

Didn't find what you were looking for?