MHB Differential equations and initial value problems

[ineedhelpnow]

did i do these two questions right?

#1
$y'=xe^{- \sin\left({x}\right)}-y \cos\left({x}\right)$

$y'+y \cos\left({x}\right) = xe^{-\sin\left({x}\right)}$$I(x)=e^{\int \ \cos\left({x}\right)dx}$

$\int \ \cos\left({x}\right)dx=\sin\left({x}\right)=e^{\sin\left({x}\right)}$$y'e^{\sin\left({x}\right)}+y \cos\left({x}\right)*e^{\sin\left({x}\right)}=xe^{-\sin\left({x}\right)}*e^{\sin\left({x}\right)}$

$y'e^{\sin\left({x}\right)}+y \cos\left({x}\right)*e^{\sin\left({x}\right)}=x$

$(ye^{\sin\left({x}\right)})'=x$

$ye^{\sin\left({x}\right)}=\int \ x dx$

$ye^{\sin\left({x}\right)}= \frac{x^2}{2}+C$

$y= \frac{x^2e ^{\sin\left({x}\right)}}{2}+Ce^{-\sin\left({x}\right)}$
#2
$\frac{du}{dt}=\frac{2t+sec^2t}{2u}$ and $u(0)=4$

$2u du = 2t+ sec^2t dt$

$\int \ 2u du=\int \ (2t+ sec^2t) dt$

$\frac{2u^3}{3}=t^2+\tan\left({t}\right)+C$

$u=\sqrt[3]{\frac{3(t^2+\tan\left({t}\right)+C)}{2}}$

now $u(0)=4:$

$4=(\frac{3(0^2+\tan\left({0}\right)+C)}{2})^{1/3}$

$4=(\frac{3C}{2})^{1/3}$

$4^3=\frac{3C}{2}$

$128=3C$

$C=\frac{128}{3}$$u=(\frac{3(t^2+\tan\left({t}\right)+\frac{128}{3})}{2})^{1/3}$

 

[evinda]

did i do these two questions right?

#1
$y'=xe^{- \sin\left({x}\right)}-y \cos\left({x}\right)$

$y'+y \cos\left({x}\right) = xe^{-\sin\left({x}\right)}$$I(x)=e^{\int \ \cos\left({x}\right)dx}$

$\int \ \cos\left({x}\right)dx=\sin\left({x}\right)=e^{\sin\left({x}\right)}$$y'e^{\sin\left({x}\right)}+y \cos\left({x}\right)*e^{\sin\left({x}\right)}=xe^{-\sin\left({x}\right)}*e^{\sin\left({x}\right)}$

$y'e^{\sin\left({x}\right)}+y \cos\left({x}\right)*e^{\sin\left({x}\right)}=x$

$(ye^{\sin\left({x}\right)})'=x$

$ye^{\sin\left({x}\right)}=\int \ x dx$

$ye^{\sin\left({x}\right)}= \frac{x^2}{2}+C$

$y= \frac{x^2e ^{\sin\left({x}\right)}}{2}+Ce^{-\sin\left({x}\right)}$

It is right,you just forgot a minus sign..The solution is :
$$y=\frac{x^2e ^{-\sin\left({x}\right)}}{2}+Ce^{-\sin\left({x}\right)}$$

#2
$\frac{du}{dt}=\frac{2t+sec^2t}{2u}$ and $u(0)=4$

$2u du = 2t+ sec^2t dt$

$\int \ 2u du=\int \ (2t+ sec^2t) dt$

$\frac{2u^3}{3}=t^2+\tan\left({t}\right)+C$

From $\int \ 2u du=\int \ (2t+ sec^2t) dt$ we get:

$$u^2=t^2+ \tan t +C$$

 

[ineedhelpnow]

thanks, its always the little things that slip past me and i don't even now how i messed up like that on the second one . so for the second one:

$u=\sqrt{t^2+ \tan\left({t}\right)+C}$

$u(0)=4$, so

$4=\sqrt{0^2+ \tan\left({0}\right)+C}$

$4=\sqrt{C}$

$C=16$

$u=\sqrt{t^2+ \tan\left({t}\right)+16}$

 

[evinda]

thanks, its always the little things that slip past me and i don't even now how i messed up like that on the second one . so for the second one:

$u=\sqrt{t^2+ \tan\left({t}\right)+C}$

$u(0)=4$, so

$4=\sqrt{0^2+ \tan\left({0}\right)+C}$

$4=\sqrt{C}$

$C=16$

$u=\sqrt{t^2+ \tan\left({t}\right)+16}$

It is :
$$u^2=t^2+ \tan t+C \Rightarrow
u= \pm \sqrt{t^2+ \tan t+C}$$

but as $u(0)=4$, we reject the negative solution,so it is as you have mentioned.. (Smile)

 

[MarkFL]

Here's an alternate way of dealing with the constant of integration you may find useful...

Suppose you have the initial value problem (IVP):

$$\frac{dy}{dx}=f(x)$$ where $$y\left(x_0\right)=y_0$$

Now, separating variables and using indefinite integrals, we may write:

$$\int\,dy=\int f(x)\,dx$$

And upon integrating, we find

$$y(x)=F(x)+C$$ where $$\frac{d}{dx}\left(F(x) \right)=f(x)$$

Using the initial condition, we get

$$y\left(x_0 \right)=F\left(x_0 \right)+C$$

Solving for $C$ and using $$y\left(x_0\right)=y_0$$, we obtain:

$$C=y_0-F\left(x_0 \right)$$ thus:

$$y(x)=F(x)+y_0-F\left(x_0 \right)$$

which we may rewrite as:

$$y(x)-y_0=F(x)-F\left(x_0 \right)$$

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

$$\int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx$$

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

$$\int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv$$

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.