MAT244--2019F > Term Test 1

Problem 1 (main sitting)

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Victor Ivrii:
(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
\bigl(y +3 y^2e^{2x}\bigr) + \bigl(1+2ye^{2x}\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(0)=1$.

Gavrilo Milanov Dzombeta:
$$M = y + 3y^2 e^{2x} \implies M_y = 1 + 6y e^{2x}$$
$$N = 1 + 2y e^{2x} \implies N_x = 4y e^{2x}$$
$$M_y \neq N_x \implies \text{ not exact}$$
$$\dfrac{M_y - N_x}{N} = 1$$
$$\mu = e^{\int 1 dx} = e^x$$
$$\left(e^x y + 3y^2 e^{3x}\right) + \left(e^x + 2y e^{3x}\right) y^\prime = 0$$
$$\tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\tilde{M_y} = e^x + 6y e^{3x}$$
$$\tilde{N} = e^x + 2y e^{3x}$$
$$\tilde{N_x} = e^x + 6y e^{3x}$$
$$\tilde{M_y} = \tilde{N_x} \implies \text{ exact equation}$$
$$\psi_x = \tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\int \psi_x dx = \int \left(e^x y + 3y^2 e^{3x}\right) dx$$
$$\therefore \psi = e^x y + y^2 e^{3x} + h\left(y\right)$$
$$\psi_y = e^x + 2y e^{3x} + h^{\prime}\left(y\right) = \tilde{N}$$
$$\therefore h^{\prime}\left(y\right) = 0 \implies h\left(y\right) = c$$
$$\therefore \psi = e^x y + y^2 e^{3x} = c$$
$$y\left(0\right) = 1 \implies 1 + 1 = c$$
$$\therefore e^x y + y^2 e^{3x} = 2$$

There is no reason to post after this, unless you disagree with this solution. Especially no point to post attachments. V.I.
Instead of sequence single equations it would be better to use multiline environment like gather or gather* to avoid excessive vertical spacing
--- Code: ---\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather}
--- End code ---

Jiaqi Huang:
The solution is typed in this pdf.

Linjie Wang:
Somehow the typed latex version can not be pasted to the forum, here are the screenshots of it ;D

BJM:
Here is the solution.

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