Very Trick laplace Transform Q

[matt222]

Homework Statement

solve the following differential equation using laplace transform:

dy/dt+yx=0
y=20 when x=0

Homework Equations

The Attempt at a Solution

I took the laplace for each term
L(dy/dx)= s*Y(s)-y(0)

L(xy)=X(s)Y(s)

subtitute back to the equation,
s*Y(s)-y(0)+X(s)Y(s)=0

s*Y(s)-20+X(s)Y(s)=0

Y(s)=20/(s+X(s))

I got until here aand in point of view it will not be solved
<is it what I did and what I said is right

same with cos(y)*dy/dt-1/t=0, y=pi/4 when t=1 it wony be solved using laplace, is it true

 

[gabbagabbahey]

Homework Statement

solve the following differential equation using laplace transform:

dy/dt+yx=0
y=20 when x=0

Do you mean \frac{dy}{dx}+xy=0[/itex] ?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>The Attempt at a Solution</h2><br /> <br /> I took the laplace for each term<br /> L(dy/dx)= s*Y(s)-y(0) </div> </div> </blockquote><br /> Assuming you are transforming from the x-domain to the s-domain, that is correct.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> L(xy)=X(s)Y(s) </div> </div> </blockquote><br /> No, that&#039;s not how you take the Laplace transform of a product of two functions.<br /> <br /> \mathcal{L}\left[x^n y(x)\right] = (-1)^n \frac{d^n}{ds^n}Y(s)

 

[matt222]

L(dy/dx)= s*Y(s)-y(0)

L(xy)=-dY(s)/ds

so now we have s*Y(s)-y(0)-dY(s)/ds=0

so now we have dY(s)/ds-s*Y(s)=-20

now by using integration factor
assume
p=-s
Q=-20

uY(s)=intgeration(u*Q)

u=exp(-s^2/2)

so u* Y(s)=integration(exp(-s^2/2)*-20))

Y(s)=40+exp(s^2/2)*c

back to x domain it will be really hard