87631
The particular solution of the differential equation \(\frac{d y}{d x}=\sec y, y(0)=0\) is
1 \(x=\cos y\)
2 \(x=\sin y+q\)
3 \(y=\sin x\)
4 \(x=\sin y\)
Explanation:
(D) : Given, differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{y}\) \(\int \cos y d y=\int d x\) \(\sin y=x+c\) \(y(0)=0\) \(c=0\) Therefore, the required solution of differential equation, \(\mathrm{x}=\sin \mathrm{y}\)
AP EAMCET-18.09.2020
Differential Equation
87632
Let \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\) and \(f(2)=5\). Using Simpson's rule, \(\int_{0}^{2} f(x) d x\) is equal to
1 \(\frac{14}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{14}{9}\)
4 \(\frac{7}{9}\)
Explanation:
(A) : Let, \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\), and \(\mathrm{f}(2)=5\), \(\because \quad \mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{4}=\frac{2}{4}=\frac{1}{2}=0.5\) \(\therefore\) By Simpson's rule, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{\mathrm{h}}{3}\left[\left(\mathrm{y}_{0}+\mathrm{y}_{4}\right)+4\left(\mathrm{y}_{1}+\mathrm{y}_{3}\right)+2\left(\mathrm{y}_{2}\right)\right]\) \(=\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right]\) \(=\frac{0.5}{3}[6+18+4]=\frac{0.5}{3} \times 28=\frac{14}{3}\)
Differential Equation
87636
If \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\), then \((x+y+z)\left(u_{x}+u_{y} u_{z}\right)\) is equal to
1 0
2 \(x-y+z\)
3 2
4 3
Explanation:
(D) : Given, \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\) Differentiating \(\mu\) partially with \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) respectively \(\frac{\partial u}{\partial x}=\frac{3 x^{2}-3 y z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{i}\) \(\frac{\partial u}{\partial y}=\frac{3 y^{2}-3 x z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{ii}\) \(\frac{\partial u}{\partial z}=\frac{3 z^{2}-3 x y}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{iii}\) On adding equation (i), (ii) \& (iii), we get - \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{3 \mathrm{x}^{2}+3 \mathrm{y}^{2}+3 \mathrm{z}^{2}-3 \mathrm{xy}-3 \mathrm{yz}-3 \mathrm{zx}}{\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xz}}\) On factorizing \(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\) From equation (iv) \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\frac{3\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right]}{(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+z y+x z)\right]}\) \(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\frac{3}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\) \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right) \times(\mathrm{x}+\mathrm{y}+\mathrm{z})=3\)
AP EAMCET-2013
Differential Equation
87637
The general solution of the differential equation \(100 \frac{d^{2} y}{d x^{2}}-20 \frac{d y}{d x}+y=0\) is
87631
The particular solution of the differential equation \(\frac{d y}{d x}=\sec y, y(0)=0\) is
1 \(x=\cos y\)
2 \(x=\sin y+q\)
3 \(y=\sin x\)
4 \(x=\sin y\)
Explanation:
(D) : Given, differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{y}\) \(\int \cos y d y=\int d x\) \(\sin y=x+c\) \(y(0)=0\) \(c=0\) Therefore, the required solution of differential equation, \(\mathrm{x}=\sin \mathrm{y}\)
AP EAMCET-18.09.2020
Differential Equation
87632
Let \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\) and \(f(2)=5\). Using Simpson's rule, \(\int_{0}^{2} f(x) d x\) is equal to
1 \(\frac{14}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{14}{9}\)
4 \(\frac{7}{9}\)
Explanation:
(A) : Let, \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\), and \(\mathrm{f}(2)=5\), \(\because \quad \mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{4}=\frac{2}{4}=\frac{1}{2}=0.5\) \(\therefore\) By Simpson's rule, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{\mathrm{h}}{3}\left[\left(\mathrm{y}_{0}+\mathrm{y}_{4}\right)+4\left(\mathrm{y}_{1}+\mathrm{y}_{3}\right)+2\left(\mathrm{y}_{2}\right)\right]\) \(=\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right]\) \(=\frac{0.5}{3}[6+18+4]=\frac{0.5}{3} \times 28=\frac{14}{3}\)
Differential Equation
87636
If \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\), then \((x+y+z)\left(u_{x}+u_{y} u_{z}\right)\) is equal to
1 0
2 \(x-y+z\)
3 2
4 3
Explanation:
(D) : Given, \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\) Differentiating \(\mu\) partially with \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) respectively \(\frac{\partial u}{\partial x}=\frac{3 x^{2}-3 y z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{i}\) \(\frac{\partial u}{\partial y}=\frac{3 y^{2}-3 x z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{ii}\) \(\frac{\partial u}{\partial z}=\frac{3 z^{2}-3 x y}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{iii}\) On adding equation (i), (ii) \& (iii), we get - \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{3 \mathrm{x}^{2}+3 \mathrm{y}^{2}+3 \mathrm{z}^{2}-3 \mathrm{xy}-3 \mathrm{yz}-3 \mathrm{zx}}{\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xz}}\) On factorizing \(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\) From equation (iv) \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\frac{3\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right]}{(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+z y+x z)\right]}\) \(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\frac{3}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\) \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right) \times(\mathrm{x}+\mathrm{y}+\mathrm{z})=3\)
AP EAMCET-2013
Differential Equation
87637
The general solution of the differential equation \(100 \frac{d^{2} y}{d x^{2}}-20 \frac{d y}{d x}+y=0\) is
87631
The particular solution of the differential equation \(\frac{d y}{d x}=\sec y, y(0)=0\) is
1 \(x=\cos y\)
2 \(x=\sin y+q\)
3 \(y=\sin x\)
4 \(x=\sin y\)
Explanation:
(D) : Given, differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{y}\) \(\int \cos y d y=\int d x\) \(\sin y=x+c\) \(y(0)=0\) \(c=0\) Therefore, the required solution of differential equation, \(\mathrm{x}=\sin \mathrm{y}\)
AP EAMCET-18.09.2020
Differential Equation
87632
Let \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\) and \(f(2)=5\). Using Simpson's rule, \(\int_{0}^{2} f(x) d x\) is equal to
1 \(\frac{14}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{14}{9}\)
4 \(\frac{7}{9}\)
Explanation:
(A) : Let, \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\), and \(\mathrm{f}(2)=5\), \(\because \quad \mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{4}=\frac{2}{4}=\frac{1}{2}=0.5\) \(\therefore\) By Simpson's rule, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{\mathrm{h}}{3}\left[\left(\mathrm{y}_{0}+\mathrm{y}_{4}\right)+4\left(\mathrm{y}_{1}+\mathrm{y}_{3}\right)+2\left(\mathrm{y}_{2}\right)\right]\) \(=\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right]\) \(=\frac{0.5}{3}[6+18+4]=\frac{0.5}{3} \times 28=\frac{14}{3}\)
Differential Equation
87636
If \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\), then \((x+y+z)\left(u_{x}+u_{y} u_{z}\right)\) is equal to
1 0
2 \(x-y+z\)
3 2
4 3
Explanation:
(D) : Given, \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\) Differentiating \(\mu\) partially with \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) respectively \(\frac{\partial u}{\partial x}=\frac{3 x^{2}-3 y z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{i}\) \(\frac{\partial u}{\partial y}=\frac{3 y^{2}-3 x z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{ii}\) \(\frac{\partial u}{\partial z}=\frac{3 z^{2}-3 x y}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{iii}\) On adding equation (i), (ii) \& (iii), we get - \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{3 \mathrm{x}^{2}+3 \mathrm{y}^{2}+3 \mathrm{z}^{2}-3 \mathrm{xy}-3 \mathrm{yz}-3 \mathrm{zx}}{\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xz}}\) On factorizing \(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\) From equation (iv) \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\frac{3\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right]}{(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+z y+x z)\right]}\) \(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\frac{3}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\) \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right) \times(\mathrm{x}+\mathrm{y}+\mathrm{z})=3\)
AP EAMCET-2013
Differential Equation
87637
The general solution of the differential equation \(100 \frac{d^{2} y}{d x^{2}}-20 \frac{d y}{d x}+y=0\) is
87631
The particular solution of the differential equation \(\frac{d y}{d x}=\sec y, y(0)=0\) is
1 \(x=\cos y\)
2 \(x=\sin y+q\)
3 \(y=\sin x\)
4 \(x=\sin y\)
Explanation:
(D) : Given, differential equation, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\sec \mathrm{y}\) \(\int \cos y d y=\int d x\) \(\sin y=x+c\) \(y(0)=0\) \(c=0\) Therefore, the required solution of differential equation, \(\mathrm{x}=\sin \mathrm{y}\)
AP EAMCET-18.09.2020
Differential Equation
87632
Let \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\) and \(f(2)=5\). Using Simpson's rule, \(\int_{0}^{2} f(x) d x\) is equal to
1 \(\frac{14}{3}\)
2 \(\frac{7}{6}\)
3 \(\frac{14}{9}\)
4 \(\frac{7}{9}\)
Explanation:
(A) : Let, \(\mathrm{f}(0)=1, \mathrm{f}(0.5)=\frac{5}{4}, \mathrm{f}(1)=2, \mathrm{f}(1.5)=\frac{13}{4}\), and \(\mathrm{f}(2)=5\), \(\because \quad \mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{4}=\frac{2}{4}=\frac{1}{2}=0.5\) \(\therefore\) By Simpson's rule, \(\int_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{\mathrm{h}}{3}\left[\left(\mathrm{y}_{0}+\mathrm{y}_{4}\right)+4\left(\mathrm{y}_{1}+\mathrm{y}_{3}\right)+2\left(\mathrm{y}_{2}\right)\right]\) \(=\frac{0.5}{3}\left[(1+5)+4\left(\frac{5}{4}+\frac{13}{4}\right)+2(2)\right]\) \(=\frac{0.5}{3}[6+18+4]=\frac{0.5}{3} \times 28=\frac{14}{3}\)
Differential Equation
87636
If \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\), then \((x+y+z)\left(u_{x}+u_{y} u_{z}\right)\) is equal to
1 0
2 \(x-y+z\)
3 2
4 3
Explanation:
(D) : Given, \(u=\log \left(x^{3}+y^{3}+z^{3}-3 x y z\right)\) Differentiating \(\mu\) partially with \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) respectively \(\frac{\partial u}{\partial x}=\frac{3 x^{2}-3 y z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{i}\) \(\frac{\partial u}{\partial y}=\frac{3 y^{2}-3 x z}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{ii}\) \(\frac{\partial u}{\partial z}=\frac{3 z^{2}-3 x y}{x^{3}+y^{3}+z^{3}-3 x y z} \tag{iii}\) On adding equation (i), (ii) \& (iii), we get - \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{3 \mathrm{x}^{2}+3 \mathrm{y}^{2}+3 \mathrm{z}^{2}-3 \mathrm{xy}-3 \mathrm{yz}-3 \mathrm{zx}}{\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xz}}\) On factorizing \(\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}-3 \mathrm{xyz}\) \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-(\mathrm{xy}+\mathrm{yz}+\mathrm{xz})\) From equation (iv) \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=\frac{3\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right]}{(x+y+z)\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+z y+x z)\right]}\) \(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\frac{3}{\mathrm{x}+\mathrm{y}+\mathrm{z}}\) \(\left(\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\frac{\partial \mathrm{u}}{\partial \mathrm{y}}+\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right) \times(\mathrm{x}+\mathrm{y}+\mathrm{z})=3\)
AP EAMCET-2013
Differential Equation
87637
The general solution of the differential equation \(100 \frac{d^{2} y}{d x^{2}}-20 \frac{d y}{d x}+y=0\) is
