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### Topics - yueyangyu

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##### Quiz-5 / quiz5
« on: November 01, 2019, 02:40:58 AM »
Verify  that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$$(1-t)y''+ty'-y=2(t-1)^2e^{-t}, 0<t<1;y1(t)=e^t \quad y2(t)=t$$

$$y1(t)=e^t \quad y'1(t)=e^t \quad y''1(t)=e^t$$
$$y2(t)=t \quad y'2(t)=1 \quad y''2(t)=0$$
Substitude back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
Verified that y1(t) and y2(t) both satisfy the corresponding homogeneous equation.
$$y_c(t)=C_1e^t+C_2t$$
Divide both side by 1-t, then
$$p(t)=\frac{t}{1-t} \quad q(t)=-\frac{1}{1-t} \quad g(t)=-2(t-1)e^{-t}$$
$$W(t)=(1-t)e^t$$
$$u1(t)=-\int{\frac{y2(t)g(t)}{W(t)}}dt=-2\int{te^{2t}}=(t+\frac{1}{2})e^{-2t}$$
$$u2(t)=\int\frac{y1(t)g(t)}{W(t)}dt=2\int{e^{-t}}=-2e^{-t}$$
Therefore, the particular solution is:
$$Y(t)=u1(t)y1(t)+u2(t)y2(t)=(t+\frac{1}{2})e^{-2t}*e^t+(-2e^{-t})*t=(\frac{1}{2}-t)e^{-t}$$
Hence, the general solution:
$$y(t)=y_c(t)+Y(t)=C_1e^t+C_2t+(\frac{1}{2}-t)e^{-t}$$
The particular solution is:
$$Y(t)=(\frac{1}{2}-t)e^{-t}$$

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##### Quiz-4 / quiz4 tut0502
« on: October 18, 2019, 04:05:06 PM »
Find the general solution of the differential equation
$y''+2y'+2y=0$

The characteristic equation of the given equation is:

$r^2+2r+2=0$

$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm{i}$

Then,
$$r_1=-1+i \quad r_2=-1-i$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{-t}cost+c_2e^{-t}sint$$

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##### Quiz-3 / quiz3 tut0502
« on: October 11, 2019, 02:00:02 PM »
Find the general solution of the given differential equation.
$$y''-2y'-2y=0$$

Assume that $$y=e^{rt}$$ and it follows that r must be a root of characteristic equation
$$r^2-2r-2=0$$
$$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Then,
$$r_1=1+\sqrt{3} \quad r_2=1-\sqrt{3}$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

4
##### Quiz-2 / quiz2 0502
« on: October 04, 2019, 02:52:24 PM »
Find an integrating factor and solve the given equation:
$$(3x+\frac{6}{y})+(\frac{x^{2}}{y}+3\frac{y}{x})\frac{dy}{dx}=0$$

Firstly, find an integrating factor $\mu$ as a function of xy s.t.
$$(\mu M)_y=(\mu N)_x$$
Let $$z=xy$$.
$$\mu M_y+xM\frac{d\mu}{dz}=\mu N_x+yN\frac{d\mu}{dz}$$

$$\frac{d\mu}{dz}=\mu(\frac{N_x-M_y}{xM-yN})$$
Therefore,
$$\mu(z)=exp(\int(R(z))dz)$$
where $$R(z)=R(xy)=\frac{N_x-M_y}{xM-yN}$$
Let $$M=3x+\frac{6}{y}$$ $$N=\frac{x^{2}}{y}+3\frac{y}{x}$$
Then,
$$M_y=\frac{-6}{y^{2}} \quad N_x=\frac{2x}{y}-\frac{37}{x^{2}}$$
We can see that this equation is not exact
$$\frac{N_x-M_y}{xM-yN}=\frac{\frac{2x}{y}-\frac{3y}{x^{2}}+\frac{6}{y^{2}}}{2x^{2}+\frac{6x}{y}-\frac{3y^{2}}{x}}=\frac{1}{xy}$$
Thus, we have an integrating factor
$$\mu(xy)=exp(\int\frac{1}{z}dz)=z=xy$$
Multiplying the original equation by the integrating factor, we have
$$(3x^{2}y+6x)+(x^{3}+3y{2})\frac{dy}{dx}=0$$
This equation is exact because
$$M_y=N_x=3x^{2}$$
Thus, there exists a function $$\Psi(x,y)$$ such that $$\Psi_x(x,y)=3x^{2}y+6x$$
$$\Psi_y(x,y)=x^{3}+3y{2}$$
$$\Psi(x,y)=x^{3}y+3x^{2}+h(y)$$
Differentiating with respect to y, we get
$$\Psi_y(x,y)=x^{3}+h'(y)$$
$$h'(y)=3y^{2} \quad h(y)=y^{3}$$
and we have
$$\Psi(x,y)=x^{3}y+3x^{2}+y^{3}$$
Thus the solutions of the differential equation are given implicitly by
$$x^{3}y+3x^{2}+y^{3}=C$$

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