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 March 19th, 2012, 08:18 AM #1 Newbie   Joined: Mar 2012 Posts: 4 Thanks: 0 Please solve the 6 ODEs... Could you please solve the attached ODEs?
March 19th, 2012, 11:17 AM   #2
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Wow. Six problems in an attachment requiring MSWord and no work shown.

Quote:
 1. Solve the ODE $x\frac{dy}{dx}=y-\sqrt{x^2+y^2}$
Dividing through by x we obtain:

$\frac{dy}{dx}=\frac{y}{x}-\sqrt{$$\frac{y}{x}$$^2+1}$

This is a first order homogeneous equation, so let:

$v=\frac{y}{x}\:\therefore\:\frac{dy}{dx}=v+x\frac{ dv}{dx}$

With these substitutions, the ODE becomes:

$v+x\frac{dv}{dx}=v-\sqrt{v^2+1}$

$x\frac{dv}{dx}=-\sqrt{v^2+1}$

This is a separable equation, so we have:

$\frac{dv}{\sqrt{v^2+1}}=-\frac{dx}{x}$

Integrating gives:

$\sinh^{\small{-1}}(v)=C-\ln\|x\|$

$v=\sinh$$C-\ln\left|x\right|$$$

Substitute back for v:

$y(x)=x\sinh$$C-\ln\left|x\right|$$$

Quote:
 2. Solve the ODE $x\frac{dy}{dx}+y+x^2y^2=0$
If we divide through by x and arrange the ODE as:

$\frac{dy}{dx}+$$\frac{1}{x}$$y=$$-x$$y^2$

we have a Bernoulli equation. Dividing through by $y^2$ we obtain:

$y^{\small{-2}}\frac{dy}{dx}+$$\frac{1}{x}$$y^{\small{-1}}=-x$

The substitution $v=\frac{1}{y}\:\therefore\:\frac{dy}{dx}=-y^2\frac{dv}{dx}$ will give us the linear equation:

$\frac{dv}{dx}-\frac{1}{x}v=x$

Divide through by x:

$\frac{1}{x}\cdot\frac{dv}{dx}+$$-\frac{1}{x^2}$$v=1$

$\frac{d}{dx}$$\frac{v}{x}$$=1$

$\int\,d$$\frac{v}{x}$$=\int\,dx$

$\frac{v}{x}=x+C$

$v=x(x+C)$

Substitute back for v:

$\frac{1}{y}=x(x+C)$

$y=\frac{1}{x(x+C)}$ where $x\ne0$

We must observe that we lost the trivial solution $y=0$ in the above process.

Quote:
 3. Find the complementary solution $y_c$ to the homogeneous ODE $\frac{d^4y}{dx^4}-4\frac{d^3y}{dx^3}+7\frac{d^2y}{dx^2}-6\frac{dy}{dx}+2y=0$
The associated auxiliary equation is:

$r^4-4r^3+7r^2-6r+2=0$

$(r-1)^2$$r^2-2r+2$$=0$

We have the real root $r=1$ of multiplicity 2 and the complex roots $r=1\pm i$ and so we have:

$y_c(x)=c_1e^x+c_2xe^x+c_3e^x\cos(x)+c_4e^x\sin(x)$

March 19th, 2012, 06:48 PM   #3
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Re: Please solve the 6 ODEs...

Quote:
 4. Find the general solution to the homogeneous ODE $\frac{d^4y}{dx^4}+4k^4y=0$ where $k\ne0$
The roots of the associated auxiliary equation are:

$r=\pm(1\pm i)k$

Hence, the general solution is:

$y(x)=e^{kx}$$c_1\cos(kx)+c_2\sin(kx)$$+e^{-kx}$$c_3\cos(kx)+c_4\sin(kx)$$$

Quote:
 5. Use the method of undetermined coefficients to find the general solution of $\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=3e^{-x}+10\sin(x)-4x$
First we determine the solution $y_h(x)$ to the associated homogeneous equation, whose associate auxiliary equation has the roots:

$r=-1,\,2$ hence:

$y_h(x)=c_1e^{-x}+c_2e^{2x}$

Next, we determine the particular solution $y_{p_1}(x)$ associated with the term $g_1(x)=3e^{-x}$

Since $g_1(x)$ is a solution to the corresponding homogeneous equation, we assume the particular solution is of the form:

$y_{p_1}(x)=Axe^{-x}$

$y_{p_1}'(x)=A$$-xe^{-x}+e^{-x}$$=Ae^{-x}$$1-x$$$

$y){p_1}''(x)=A$$e^{-x}(-1)-(1-x)e^{-x}$$=Ae^{-x}(x-2)$

Substitution yields:

$$$Ae^{-x}(x-2)$$-$$Ae^{-x}\(1-x$$\)-2$$Axe^{-x}$$=3e^{-x}$

$A(x-2)+$$A\(x-1$$\)-2Ax=3$

$-3A=3$

$A=-1$ and so:

$y_{p_1}(x)=-xe^{-x}$

Now for the second particular solution $y_{p_2}(x)$ associated with the term $g_2(x)=10\sin(x)$.

We may assume the particular solution is of the form:

$y_{p_2}(x)=A\cos(x)+B\sin(x)$

$y_{p_2}'(x)=-A\sin(x)+B\cos(x)$

$y_{p_2}''(x)=-A\cos(x)-B\sin(x)=-y_{p_2}(x)$

Substitution yields:

$$$-y_{p_2}(x)$$-y_{p_2}'(x)-2y_{p_2}(x)=10\sin(x)$

$-y_{p_2}'(x)-3y_{p_2}(x)=10\sin(x)$

$-$$-A\sin(x)+B\cos(x)$$-3$$A\cos(x)+B\sin(x)$$=10\sin(x)$

$(A-3B)\sin(x)-(3A+B)\cos(x)=10\sin(x)-0\cos(x)$

Equating coefficients yields:

$A-3B=10$

$3A+B=0$

Solving this system yields:

$A=1,\,B=-3$ and so:

$y_{p_2}(x)=\cos(x)-3\sin(x)$

Lastly, we determine the third particular solution $y_{p_3}(x)$ associated with the term $g_3(x)=-4x$.

We may assume the particular solution is of the form:

$y_{p_3}(x)=Ax+B$

$y_{p_3}'(x)=A$

$y_{p_3}''(x)=0$

Substitution yields:

$$$0$$-$$A$$-2$$Ax+B$$=-4x$

$2Ax+(A+2B)=4x+0$

Equating coefficients yields:

$2A=4\:\therefore\:A=2$

$A+2B=0\:\therefore\:B=-1$ and so:

$y_{p_3}(x)=2x-1$

Then, the general solution is:

$y(x)=y_h(x)+y_{p_1}(x)+y_{p_2}(x)+y_{p_3}(x)=c_1e^ {-x}+c_2e^{2x}-xe^{-x}+\cos(x)-3\sin(x)+2x-1$

Quote:
 6. Consider the ODE: $x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+6=0$ By using the substitution $z=\ln(x)$, show that the given differential equation can be converted into an ordinary differential equation with constant coefficients. Solve this differential equation and hence obtain the solution of y as a function of x.
This is what's known as a Cauchy-Euler equation. The required substitution implies:

$x=e^z$ and it follows from the chain rule that:

$\frac{dy}{dz}=\frac{dy}{dx}\cdot\frac{dx}{dz}=\fra c{dy}{dx}e^z=x\frac{dy}{dx}$

$\frac{dy}{dz}=x\frac{dy}{dx}$

Differentiating with respect to z:

$\frac{d}{dz}$$\frac{dy}{dz}$$=\frac{d}{dz}$$x\frac {dy}{dx}$$$

$\frac{d^2y}{dz^2}=\frac{dx}{dz}\cdot\frac{dy}{dx}+ x\frac{d}{dz}$$\frac{dy}{dx}$$=\frac{dy}{dz}+x\fra c{d^2y}{dx^2}\cdot\frac{dx}{dz}=\frac{dy}{dz}+x^2\ frac{d^2y}{dx^2}$

Hence:

$x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dz^2}-\frac{dy}{dz}$

Substituting for $x\frac{dy}{dx}$ and $x^2\frac{d^2y}{dx^2}$ into the original ODE yields:

$$$\frac{d^2y}{dz^2}-\frac{dy}{dz}$$+4$$\frac{dy}{dz}$$+6=0$

$\frac{d^2y}{dz^2}+3\frac{dy}{dz}+6=0$

Now we have a homogeneous equation with constant coefficients, whose associated auxiliary equation has the roots:

$r=\frac{-3\pm i\sqrt{15}}{2}$

and so the general solution $y(z)$ is:

$y(z)=c_1e^{-\frac{3}{2}z}\cos$$\frac{\sqrt{15}}{2}z$$+c_2e^{-\frac{3}{2}z}\sin$$\frac{\sqrt{15}}{2}z$$$

Substituting back for z we have:

$y(x)=c_1e^{-\frac{3}{2}\ln(x)}\cos$$\frac{\sqrt{15}}{2}\ln(x)\ )+c_2e^{-\frac{3}{2}\ln(x)}\sin\(\frac{\sqrt{15}}{2}\ln(x)\ )$ $y(x)=x^{\small{-\frac{3}{2}}}\(c_1\cos\(\frac{\sqrt{15}}{2}\ln(x)\ )+c_2\sin\(\frac{\sqrt{15}}{2}\ln(x)$$\)$

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