This set of Engineering Mathematics Question Bank focuses on “Euler’s Theorem – 2”.
1. In euler theorem x ∂z⁄∂x + y ∂z⁄∂y = nz, here ‘n’ indicates?
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z
View Answer
Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”.
2. If z = xn f(y⁄x) then?
a) y ∂z⁄∂x + x ∂z⁄∂y = nz
b) 1/y ∂z⁄∂x + 1/x ∂z⁄∂y = nz
c) x ∂z⁄∂x + y ∂z⁄∂y = nz
d) 1/x ∂z⁄∂x + 1/y ∂z⁄∂y = nz
View Answer
Explanation: Since the given function is homogeneous of order n, hence by euler’s theorem
x ∂z⁄∂x + y ∂z⁄∂y = nz.
3. Necessary condition of euler’s theorem is?
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
View Answer
Explanation:
Answer ‘z should be homogeneous and of order n’ is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”
Answer ‘z should not be homogeneous but of order n’ is incorrect as z should be homogeneous.
Answer ‘z should be implicit’ is incorrect as z should not be implicit.
Answer ‘z should be the function of x and y only’ is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.
4. If \(z=e^{\frac{x^2+y^2}{x+y}}\) then, \(x \frac{∂z}{∂x} + y \frac{∂z}{∂y}\) is?
a) 0
b) zln(z)
c) z2 ln(z)
d) z
View Answer
Explanation:
Given \(z=e^{\frac{x^2+y^2}{x+y}}\),let u=ln(z)=\(\frac{x^2+y^2}{x+y}=\frac{x(1+(\frac{y}{x})^2)}{(1+\frac{y}{x})}\) = x f(y/x)
Hence u is homogeneous of order 1,
Hence,
\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}\)=u
Putting, u = ln(z) we get,
\(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\) = zln(z)
5. If \(z=sin^{-1}\frac{x^3+y^3+z^3}{x+y+z}\) then, \(x\frac{∂z}{∂x}+y\frac{∂z}{∂y}\).
a) 2 tan(z)
b) 2 cot(z)
c) tan(z)
d) cot(z)
View Answer
Explanation:
Given \(z=sin^{-1}\frac{x^3+y^3+z^3}{x+y+z}\), put u=sin(z)=\(\frac{x^3+y^3+z^3}{x+y+z}=x^2 f(\frac{y}{x},\frac{z}{x})\)
Hence, \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=2u\)
Putting u = sin(z), we get
\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=\frac{2Sin(z)}{Cos(z)}=2Tan(z)\)
6. Value of \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}\) if \(u=\frac{Sin^{-1} (\frac{y}{x})(\sqrt{x}+\sqrt{y})}{x^3+y^3}\) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
View Answer
Explanation: Since the function can be written as,
u=\(x^{\frac{-5}{2}} \frac{Sin^{-1} (\frac{y}{x})(1+\sqrt{\frac{y}{x}})}{1+(\frac{y}{x})^3}=x^n f(\frac{y}{x})\), by euler’s theorem,
\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y} = -\frac{5}{2} u\)
7. If f1(x,y) and f2(x,y) are homogeneous and of order ‘n’then the function f3(x,y) = f1(x,y) + f2(x,y) satisfies euler’s theorem.
a) True
b) False
View Answer
Explanation: Since f1(x,y) and f2(x,y) are homogeneous and of order n hence,
\(x \frac{∂f_1}{∂x}+y \frac{∂f_1}{∂y} = nf_1 (x,y)\)
\(x \frac{∂f_2}{∂x}+y \frac{∂f_2}{∂y} = nf_2 (x,y)\)
Hence adding these two equations,
We get
\(x \frac{∂f_1+f_2}{∂x}+y \frac{∂f_1+f_2}{∂y} = nf_2 (x,y)+nf_1 (x,y)\)
\(x \frac{∂f_3}{∂x}+y \frac{∂f_3}{∂y} = nf_3 (x,y)\)
Hence f3 satisfies euler’s theorem.
8. If \(z=ln(\frac{x^2+y^2}{x+y})-e^{\frac{x^2+y^2}{x+y}}\) then find \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}\).
a) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
b) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
c) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
d) \(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
View Answer
Explanation:
Given \(z=ln(\frac{x^2+y^2}{x+y})-e^\frac{x^2+y^2}{x+y}\)
Let, \(u = ln(\frac{x^2+y^2}{x+y})\) and \(v=e^(\frac{x^2+y^2}{x+y})\) hence z=u-v
Now, let \(u’ = e^u = \frac{x^2+y^2}{x+y}=xf(\frac{y}{x})\) hence u’ satisfies euler’s theorem,
Hence,
\(x \frac{∂u’}{∂x}+y \frac{∂u’}{∂y}=u’\)
Hence, by putting u’=eu, we get
\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=\frac{e^u}{e^u} = 1\), ……(1)
Now, let v’ = ln(v)= \(\frac{x^2+y^2}{x+y}=xf(\frac{y}{x})\) hence v’ satisfies euler’s theorem,
Hence,
\(x \frac{∂v’}{∂x}+y \frac{∂v’}{∂y}=v’\)
Hence, by putting v’=ln(v),we get
\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=vln(v)\), …… (2)
By subtracting eq(1) and eq(2), we get
\(x \frac{∂z}{∂x}-y \frac{∂z}{∂y}=1-\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
9. If z = Sin-1 (x⁄y) + Tan-1 (y⁄x) then x ∂z⁄∂x + y ∂z⁄∂y is?
a) 0
b) y
c) 1 + x⁄y Sin-1 (x⁄y)
d) 1 + y⁄x Tan-1 (y⁄x)
View Answer
Explanation: Given z = Sin-1 (x⁄y) + Tan-1 (y⁄x)
Let, u = Sin-1 (x⁄y) and v = Tan-1 (y⁄x) hence z = u + v
Now, let u’ = Sin(u) = x⁄y = f(x⁄y) hence u’ satisfies euler’s theorem,
Hence,
\(x \frac{∂u’}{∂x}+y \frac{∂u’}{∂y}=0\)
Hence, by putting u’=eu, we get
\(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=0/e^u = 0\) ,……(1)
Now, let v’= Tan(v)=y/x=f(y/x) hence v’ satisfies euler’s theorem,
Hence,
\(x \frac{∂v’}{∂x}+y \frac{∂v’}{∂y}=0\)
Hence, by putting v’=ln(v), we get
\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=0 \),……(2)
By adding eq(1) and eq(2), we get
\(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=1+\frac{x^2+y^2}{x+y} e^{\frac{x^2+y^2}{x+y}}\)
10. If f(x,y)is a function satisfying euler’ s theorem then?
a) \(x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
b) \(\frac{1}{x}^2 \frac{∂^2 f}{∂x^2}+2/xy \frac{∂^2 f}{∂x∂y}+\frac{1}{y}^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
c) \(x^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+y^2 \frac{∂^2 f}{∂y^2}=nf\)
d) \(y^2 \frac{∂^2 f}{∂x^2}+2xy \frac{∂^2 f}{∂x∂y}+x^2 \frac{∂^2 f}{∂y^2}=n(n-1)f\)
View Answer
Explanation: Since f satisfies euler’s theorem,
\(x \frac{∂z}{∂x}+y \frac{∂z}{∂y}=nz\)
Differentiating it w.r.t x and y respectively we get,
\(x \frac{∂^2 u}{∂x^2}+\frac{∂u}{∂x}+y \frac{∂^2 u}{∂x∂y}=n \frac{∂u}{∂x}\),
and
\(x \frac{∂^2 u}{∂y}∂x+\frac{∂u}{∂y}+y \frac{∂^2 u}{∂y^2}=n \frac{∂u}{∂y}\)
Multiplying with x and y respectively,
\(x^2 \frac{∂^2 u}{∂x^2}+x \frac{∂u}{∂x}+xy \frac{∂^2 u}{∂x∂y}=nx \frac{∂u}{∂x}\),
and
\(xy \frac{∂^2 u}{∂y}∂x+y \frac{∂u}{∂y}+y^2 \frac{∂^2 u}{∂y^2}=ny \frac{∂u}{∂y}\)
Adding above equations we get
\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}=n(n-1)u\)
11. If \(u = Tan^{-1} (\frac{x^3+y^3}{x+y})\) then, \(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) is?
a) Sin(4u) – Cos(2u)
b) Sin(4u) – Sin(2u)
c) Cos(4u) – Sin(2u)
d) Cos(4u) – Cos(2u)
View Answer
Explanation:
Let, v = Tan(u) = x2 f(y/x)
By euler’s theorem,
g(u) = \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y} = 2\frac{Tan(u)}{Sec^2 (u)} = Sin(2u)\)
Hence,
\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y} = g(u)[g’(u)-1]\)
\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = Sin(2u)[2Cos(2u)-1] = Sin(4u)-Sin(2u)
12. If \(u = e^{\frac{(x^2+y^2)}{x+y}}\) Then, \(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\)=?
a) u ln(u)
b) u ln(u)2
c) u [1+ln(u)]
d) 0
View Answer
Explanation: Let, v = ln(u) = \(\frac{x^2+y^2}{x+y} = x f(\frac{y}{x})\)
Hence by applying euler theorem,
\(x \frac{∂v}{∂x}+y \frac{∂v}{∂y}=v\)
Hence,
g(u) = \(x \frac{∂u}{∂x}+y \frac{∂u}{∂y}=u ln(u)\)
Hence,
\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = g(u)[g’(u)-1]
\(x^2 \frac{∂^2 u}{∂x^2}+y^2 \frac{∂^2 u}{∂y}+2xy \frac{∂^2 u}{∂x∂y}\) = u ln(u)[1+ln(u)-1] = u ln(u)2
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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