MAT244--2019F > Term Test 1

Problem 1 (main sitting)

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yueyangyu:
$$M_{y}=1+6ye^{3x}$$
$$N_{x}=4ye^{2x}$$
$M_{y} ≠N_{x}$,it is not exact
$$R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x} }{1+2ye^{2x}}=1$$
$$μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x$$
Multiplying both sides by $\mu$, we get
$$ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0$$
$$M_{y}^\prime=e^x+6ye^{3x}$$
$$N_{x}^\prime=e^x+6ye^{3x}$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime$$
$$φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)$$
$$φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}$$
$h(y)^\prime=0$

So h(y)=c
$$φ(x,y)=ye^x+y^{2}e^{3x} =c$$
Since y(0)=1
$$1⋅e^0+1^2⋅e^0=2=c$$
$$φ(x,y)=ye^x+y^{2}e^{3x} =2$$

Xinyu Jing:
𝑀𝑦=$1+6𝑦𝑒^{3𝑥}$

𝑁𝑥=$4𝑦𝑒^{2𝑥}$

𝑀𝑦≠𝑁𝑥,it is not exact

$𝑅_{2}$=(𝑀𝑦−𝑁𝑥)/𝑁=$\frac{1+2𝑦𝑒^{2𝑥}}{1+2𝑦𝑒^{2𝑥}}$=1

μ=$𝑒∫𝑅_{2}𝑑𝑥$=𝑒∫1𝑑𝑥=$𝑒^{𝑥}$

Multiplying both sides by 𝜇, we get

$𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}+(𝑒𝑥+2𝑦𝑒^{3𝑥})𝑦′=0$

$𝑀′𝑦=𝑒𝑥+6𝑦𝑒^{3𝑥}$

$𝑁′𝑥=𝑒𝑥+6𝑦𝑒^{3𝑥}$

𝑀′𝑦=𝑁′𝑥,it is exact

∃φ(𝑥,𝑦)𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 φ𝑥=𝑀′,φ𝑦=𝑁′

φ(𝑥,𝑦)=∫𝑀′𝑑𝑥=$∫𝑦𝑒^{𝑥}+3𝑦_{2}𝑒^{3𝑥}𝑑𝑥=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}+ℎ(𝑦)$

$φ𝑦=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}+ℎ(𝑦)′=𝑒^{𝑥}+2𝑦𝑒^{3𝑥}$

Then ℎ(𝑦)′=0

Hence h(y)=c

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=𝑐$

Since y(0)=1

$1⋅𝑒^{0}+12⋅𝑒^{0}=2=𝑐$

$φ(𝑥,𝑦)=𝑦𝑒^{𝑥}+𝑦_{2}𝑒^{3𝑥}=2$