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Natali [406]
4 years ago
11

Estimate the figure to the nearest whole number

Mathematics
1 answer:
olya-2409 [2.1K]4 years ago
3 0
That means that, for example:

 if your number is 49 the nearest whole number would be 50.

so by saying estimate the figure to the nearest whole number , they are basically saying find your nearest whole number.

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Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
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Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

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Need help with linear function
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y = 1/2x + 2

Step-by-step explanation:

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