# Thread: Separable Ordinary D.E (help verify my working)

1. ## Separable Ordinary D.E (help verify my working)

Hi everyone

Need help to verify my working, thank you in advance for all help & support.

Solve the following separable ordinary D.E.

a) $\displaystyle \frac{dy}{dx}=(5y-2)sinx$
$\displaystyle \frac{dy}{5y-2}=sinxdx$
$\displaystyle \int\frac{dy}{5y-2}=\int sinxdx+c$
$\displaystyle \frac{1}{5}ln(5y-2)=-cosx+c$

b) $\displaystyle \frac{dy}{dx}=\frac{x^2}{y+4}$
$\displaystyle dy(y+4)=dx(x^2)$
$\displaystyle \int (y+4)dy=\int (x^2)dx+c$
$\displaystyle \frac{y^2}{2}+4y=\frac{x^3}{3}+c$

c)$\displaystyle \frac{dy}{dx}=e^{x+y}$
$\displaystyle \frac{dy}{dx}=e^xe^y$
$\displaystyle e^{-y}dy=e^{x}dx$
$\displaystyle \int e^{-y}dy=\int e^{x}dx+c$
$\displaystyle -e^{-y}=e^x+c$

d) $\displaystyle xy\frac{dy}{dx}=x^2+y^2$,Hint:substitute $\displaystyle u=\frac{y}{x}$
$\displaystyle \frac{dy}{dx}=\frac{x}{y}+\frac{y}{x}$
$\displaystyle u=\frac{y}{x}$,y=ux
$\displaystyle \frac{dy}{dx}=u+x\frac{du}{dx}$
$\displaystyle u+x\frac{du}{dx}=\frac{1}{u}+u$
$\displaystyle udu=\frac{1}{x}dx$
$\displaystyle \int udu=\int \frac{1}{x}dx+c$
$\displaystyle \frac{u^2}{2}=ln|x|+c$
$\displaystyle \frac{y^2}{2x^2}=ln|x|+c$

2. Originally Posted by anderson
Hi everyone

Need help to verify my working, thank you in advance for all help & support.

Solve the following separable ordinary D.E.

a) $\displaystyle \frac{dy}{dx}=(5y-2)sinx$
$\displaystyle \frac{dy}{5y-2}=sinxdx$
$\displaystyle \int\frac{dy}{5y-2}=\int sinxdx+c$
$\displaystyle \frac{1}{5}{\color{red}ln(5y-2)}=-cosx+c$

b) $\displaystyle \frac{dy}{dx}=\frac{x^2}{y+4}$
$\displaystyle dy(y+4)=dx(x^2)$
$\displaystyle \int (y+4)dy=\int (x^2)dx+c$
$\displaystyle \frac{y^2}{2}+4y=\frac{x^3}{3}+c$

c)$\displaystyle \frac{dy}{dx}=e^{x+y}$
$\displaystyle \frac{dy}{dx}=e^xe^y$
$\displaystyle e^{-y}dy=e^{x}dx$
$\displaystyle \int e^{-y}dy=\int e^{x}dx+c$
$\displaystyle -e^{-y}=e^x+c$

d) $\displaystyle xy\frac{dy}{dx}=x^2+y^2$,Hint:substitute $\displaystyle u=\frac{y}{x}$
$\displaystyle \frac{dy}{dx}=\frac{x}{y}+\frac{y}{x}$
$\displaystyle u=\frac{y}{x}$,y=ux
$\displaystyle \frac{dy}{dx}=u+x\frac{du}{dx}$
$\displaystyle u+x\frac{du}{dx}=\frac{1}{u}+u$
$\displaystyle udu=\frac{1}{x}dx$
$\displaystyle \int udu=\int \frac{1}{x}dx+c$
$\displaystyle \frac{u^2}{2}=ln|x|+c$
$\displaystyle \frac{y^2}{2x^2}=ln|x|+c$

Dear anderson,

In the first problem (which I have highlighted) you have'nt written the modulus sign for the natural logarithm. Other than this all the other solutions are perfect.

3. Originally Posted by anderson
Hi everyone

Need help to verify my working, thank you in advance for all help & support.

Solve the following separable ordinary D.E.

a) $\displaystyle \frac{dy}{dx}=(5y-2)sinx$
$\displaystyle \frac{dy}{5y-2}=sinxdx$
$\displaystyle \int\frac{dy}{5y-2}=\int sinxdx+c$.................................................. (a)
$\displaystyle \frac{1}{5}ln(5y-2)=-cosx+c$

b) $\displaystyle \frac{dy}{dx}=\frac{x^2}{y+4}$
$\displaystyle dy(y+4)=dx(x^2)$
$\displaystyle \int (y+4)dy=\int (x^2)dx+c$...................................(b)
$\displaystyle \frac{y^2}{2}+4y=\frac{x^3}{3}+c$

c)$\displaystyle \frac{dy}{dx}=e^{x+y}$
$\displaystyle \frac{dy}{dx}=e^xe^y$
$\displaystyle e^{-y}dy=e^{x}dx$
$\displaystyle \int e^{-y}dy=\int e^{x}dx+c$.................................................. ........(c)
$\displaystyle -e^{-y}=e^x+c$

d) $\displaystyle xy\frac{dy}{dx}=x^2+y^2$,Hint:substitute $\displaystyle u=\frac{y}{x}$
$\displaystyle \frac{dy}{dx}=\frac{x}{y}+\frac{y}{x}$
$\displaystyle u=\frac{y}{x}$,y=ux
$\displaystyle \frac{dy}{dx}=u+x\frac{du}{dx}$
$\displaystyle u+x\frac{du}{dx}=\frac{1}{u}+u$
$\displaystyle udu=\frac{1}{x}dx$
$\displaystyle \int udu=\int \frac{1}{x}dx+c$.............................................(d)
$\displaystyle \frac{u^2}{2}=ln|x|+c$
$\displaystyle \frac{y^2}{2x^2}=ln|x|+c$

Remove the $\displaystyle +c$ from the lines (a),(b),(c),(d). The constant is added after you integrate, not before integration.