87151
The degree of the differential equation \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}}\) is equal to
1 2
2 3
3 4
4 \(5 / 2\)
Explanation:
(B) : Given, the differential equation- \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}} \Rightarrow \frac{d^{2} y}{d x^{2}} \cdot \sqrt{\frac{d^{2} y}{d x^{2}}}=5 y+\frac{d y}{d x}\) On squaring both side, we get- \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)=\left(5 y+\frac{d y}{d x}\right)^{2}\) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\left(5 y+\frac{d y}{d x}\right)^{2}\) So, \(\quad\) degree \(=3\), order \(=1\).
J and K CET-2011
Differential Equation
87162
The order and degree of the differential equation of the family of circles of fixed radius \(r\) with centres on the \(\mathbf{y}\)-axis are respectively
1 2,2
2 2,3
3 1,1
4 3,1
5 1,2
Explanation:
(E) : The equation of family of circles of fixed radius ' \(\mathrm{r}\) ' with centres on the \(\mathrm{y}\)-axis is \((x-0)^{2}+(y-a)^{2}=r^{2} \tag{i}\) On differentiating with respect to \(\mathrm{x}\), we have \(2 x+2(y-a) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y-a}\) \((y-a)=\frac{-x}{\left(\frac{d y}{d x}\right)}\) On putting value in equation (i), we have- \(\mathrm{x}^{2}+\frac{\mathrm{x}^{2}}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=\mathrm{r}^{2} \Rightarrow \mathrm{x}^{2}\left\{1+\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\right\}=\mathrm{r}^{2}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\) Hence, order \(=1\) and degree \(=2\)
Kerala CEE-2013
Differential Equation
87163
The order and degree of the differential equation \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) are respectively
1 3 and 1
2 3 and 3
3 1 and 3
4 3 and 2
5 2 and 2
Explanation:
(A) : Given that, \(\left[\frac{d^{3} y}{d x^{3}}\right]^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) On cubing both sides, we get \(\left[\frac{\mathrm{d}^{3} y}{\mathrm{dx}^{3}}\right]=\left[2 \frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}+\sqrt[3]{\cos ^{2} \mathrm{x}}\right]^{3}\) Hence, highest order derivation \(=3\) and degree \(=1\)
Kerala CEE-2012
Differential Equation
87164
The degree and order of the differential equation \(\mathbf{y}=\mathbf{p x}+\mathbf{x} \sqrt[3]{\mathbf{a}^{2} \mathbf{p}^{2}+\mathbf{b}^{2}}\), where \(p=\frac{\mathbf{d y}}{\mathbf{d x}}\), are respectively
1 3,1
2 1,3
3 1,1
4 3,3
5 3,2
Explanation:
(B) : Given that, \(y=p x+x \sqrt[3]{a^{2} p^{2}+b^{2}}\) Where, \(p=\frac{d y}{d x}\) \(\text { Now, } y=\frac{d y}{d x} x+x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(y-\frac{d y}{d x} x=x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(\quad\left[y-\frac{d y}{d x} x\right]^{3}=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(y^{3}-3 y^{2} \frac{d y}{d x}+3 y\left[\frac{d y}{d x}\right]^{2}-x^{3}\left[\frac{d y}{d x}\right]^{3}\) \(=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(\therefore \quad(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\) Hence, highest order of derivative \(=1\) The power of the highest order derivative \(=3\)
87151
The degree of the differential equation \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}}\) is equal to
1 2
2 3
3 4
4 \(5 / 2\)
Explanation:
(B) : Given, the differential equation- \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}} \Rightarrow \frac{d^{2} y}{d x^{2}} \cdot \sqrt{\frac{d^{2} y}{d x^{2}}}=5 y+\frac{d y}{d x}\) On squaring both side, we get- \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)=\left(5 y+\frac{d y}{d x}\right)^{2}\) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\left(5 y+\frac{d y}{d x}\right)^{2}\) So, \(\quad\) degree \(=3\), order \(=1\).
J and K CET-2011
Differential Equation
87162
The order and degree of the differential equation of the family of circles of fixed radius \(r\) with centres on the \(\mathbf{y}\)-axis are respectively
1 2,2
2 2,3
3 1,1
4 3,1
5 1,2
Explanation:
(E) : The equation of family of circles of fixed radius ' \(\mathrm{r}\) ' with centres on the \(\mathrm{y}\)-axis is \((x-0)^{2}+(y-a)^{2}=r^{2} \tag{i}\) On differentiating with respect to \(\mathrm{x}\), we have \(2 x+2(y-a) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y-a}\) \((y-a)=\frac{-x}{\left(\frac{d y}{d x}\right)}\) On putting value in equation (i), we have- \(\mathrm{x}^{2}+\frac{\mathrm{x}^{2}}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=\mathrm{r}^{2} \Rightarrow \mathrm{x}^{2}\left\{1+\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\right\}=\mathrm{r}^{2}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\) Hence, order \(=1\) and degree \(=2\)
Kerala CEE-2013
Differential Equation
87163
The order and degree of the differential equation \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) are respectively
1 3 and 1
2 3 and 3
3 1 and 3
4 3 and 2
5 2 and 2
Explanation:
(A) : Given that, \(\left[\frac{d^{3} y}{d x^{3}}\right]^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) On cubing both sides, we get \(\left[\frac{\mathrm{d}^{3} y}{\mathrm{dx}^{3}}\right]=\left[2 \frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}+\sqrt[3]{\cos ^{2} \mathrm{x}}\right]^{3}\) Hence, highest order derivation \(=3\) and degree \(=1\)
Kerala CEE-2012
Differential Equation
87164
The degree and order of the differential equation \(\mathbf{y}=\mathbf{p x}+\mathbf{x} \sqrt[3]{\mathbf{a}^{2} \mathbf{p}^{2}+\mathbf{b}^{2}}\), where \(p=\frac{\mathbf{d y}}{\mathbf{d x}}\), are respectively
1 3,1
2 1,3
3 1,1
4 3,3
5 3,2
Explanation:
(B) : Given that, \(y=p x+x \sqrt[3]{a^{2} p^{2}+b^{2}}\) Where, \(p=\frac{d y}{d x}\) \(\text { Now, } y=\frac{d y}{d x} x+x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(y-\frac{d y}{d x} x=x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(\quad\left[y-\frac{d y}{d x} x\right]^{3}=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(y^{3}-3 y^{2} \frac{d y}{d x}+3 y\left[\frac{d y}{d x}\right]^{2}-x^{3}\left[\frac{d y}{d x}\right]^{3}\) \(=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(\therefore \quad(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\) Hence, highest order of derivative \(=1\) The power of the highest order derivative \(=3\)
87151
The degree of the differential equation \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}}\) is equal to
1 2
2 3
3 4
4 \(5 / 2\)
Explanation:
(B) : Given, the differential equation- \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}} \Rightarrow \frac{d^{2} y}{d x^{2}} \cdot \sqrt{\frac{d^{2} y}{d x^{2}}}=5 y+\frac{d y}{d x}\) On squaring both side, we get- \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)=\left(5 y+\frac{d y}{d x}\right)^{2}\) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\left(5 y+\frac{d y}{d x}\right)^{2}\) So, \(\quad\) degree \(=3\), order \(=1\).
J and K CET-2011
Differential Equation
87162
The order and degree of the differential equation of the family of circles of fixed radius \(r\) with centres on the \(\mathbf{y}\)-axis are respectively
1 2,2
2 2,3
3 1,1
4 3,1
5 1,2
Explanation:
(E) : The equation of family of circles of fixed radius ' \(\mathrm{r}\) ' with centres on the \(\mathrm{y}\)-axis is \((x-0)^{2}+(y-a)^{2}=r^{2} \tag{i}\) On differentiating with respect to \(\mathrm{x}\), we have \(2 x+2(y-a) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y-a}\) \((y-a)=\frac{-x}{\left(\frac{d y}{d x}\right)}\) On putting value in equation (i), we have- \(\mathrm{x}^{2}+\frac{\mathrm{x}^{2}}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=\mathrm{r}^{2} \Rightarrow \mathrm{x}^{2}\left\{1+\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\right\}=\mathrm{r}^{2}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\) Hence, order \(=1\) and degree \(=2\)
Kerala CEE-2013
Differential Equation
87163
The order and degree of the differential equation \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) are respectively
1 3 and 1
2 3 and 3
3 1 and 3
4 3 and 2
5 2 and 2
Explanation:
(A) : Given that, \(\left[\frac{d^{3} y}{d x^{3}}\right]^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) On cubing both sides, we get \(\left[\frac{\mathrm{d}^{3} y}{\mathrm{dx}^{3}}\right]=\left[2 \frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}+\sqrt[3]{\cos ^{2} \mathrm{x}}\right]^{3}\) Hence, highest order derivation \(=3\) and degree \(=1\)
Kerala CEE-2012
Differential Equation
87164
The degree and order of the differential equation \(\mathbf{y}=\mathbf{p x}+\mathbf{x} \sqrt[3]{\mathbf{a}^{2} \mathbf{p}^{2}+\mathbf{b}^{2}}\), where \(p=\frac{\mathbf{d y}}{\mathbf{d x}}\), are respectively
1 3,1
2 1,3
3 1,1
4 3,3
5 3,2
Explanation:
(B) : Given that, \(y=p x+x \sqrt[3]{a^{2} p^{2}+b^{2}}\) Where, \(p=\frac{d y}{d x}\) \(\text { Now, } y=\frac{d y}{d x} x+x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(y-\frac{d y}{d x} x=x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(\quad\left[y-\frac{d y}{d x} x\right]^{3}=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(y^{3}-3 y^{2} \frac{d y}{d x}+3 y\left[\frac{d y}{d x}\right]^{2}-x^{3}\left[\frac{d y}{d x}\right]^{3}\) \(=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(\therefore \quad(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\) Hence, highest order of derivative \(=1\) The power of the highest order derivative \(=3\)
87151
The degree of the differential equation \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}}\) is equal to
1 2
2 3
3 4
4 \(5 / 2\)
Explanation:
(B) : Given, the differential equation- \(\frac{d^{2} y}{d x^{2}}=\frac{5 y+\frac{d y}{d x}}{\sqrt{\frac{d^{2} y}{d x^{2}}}} \Rightarrow \frac{d^{2} y}{d x^{2}} \cdot \sqrt{\frac{d^{2} y}{d x^{2}}}=5 y+\frac{d y}{d x}\) On squaring both side, we get- \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)=\left(5 y+\frac{d y}{d x}\right)^{2}\) \(\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\left(5 y+\frac{d y}{d x}\right)^{2}\) So, \(\quad\) degree \(=3\), order \(=1\).
J and K CET-2011
Differential Equation
87162
The order and degree of the differential equation of the family of circles of fixed radius \(r\) with centres on the \(\mathbf{y}\)-axis are respectively
1 2,2
2 2,3
3 1,1
4 3,1
5 1,2
Explanation:
(E) : The equation of family of circles of fixed radius ' \(\mathrm{r}\) ' with centres on the \(\mathrm{y}\)-axis is \((x-0)^{2}+(y-a)^{2}=r^{2} \tag{i}\) On differentiating with respect to \(\mathrm{x}\), we have \(2 x+2(y-a) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y-a}\) \((y-a)=\frac{-x}{\left(\frac{d y}{d x}\right)}\) On putting value in equation (i), we have- \(\mathrm{x}^{2}+\frac{\mathrm{x}^{2}}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}}=\mathrm{r}^{2} \Rightarrow \mathrm{x}^{2}\left\{1+\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\right\}=\mathrm{r}^{2}\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]^{2}\) Hence, order \(=1\) and degree \(=2\)
Kerala CEE-2013
Differential Equation
87163
The order and degree of the differential equation \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) are respectively
1 3 and 1
2 3 and 3
3 1 and 3
4 3 and 2
5 2 and 2
Explanation:
(A) : Given that, \(\left[\frac{d^{3} y}{d x^{3}}\right]^{\frac{1}{3}}=2 \frac{d^{2} y}{d x^{2}}+\sqrt[3]{\cos ^{2} x}\) On cubing both sides, we get \(\left[\frac{\mathrm{d}^{3} y}{\mathrm{dx}^{3}}\right]=\left[2 \frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}+\sqrt[3]{\cos ^{2} \mathrm{x}}\right]^{3}\) Hence, highest order derivation \(=3\) and degree \(=1\)
Kerala CEE-2012
Differential Equation
87164
The degree and order of the differential equation \(\mathbf{y}=\mathbf{p x}+\mathbf{x} \sqrt[3]{\mathbf{a}^{2} \mathbf{p}^{2}+\mathbf{b}^{2}}\), where \(p=\frac{\mathbf{d y}}{\mathbf{d x}}\), are respectively
1 3,1
2 1,3
3 1,1
4 3,3
5 3,2
Explanation:
(B) : Given that, \(y=p x+x \sqrt[3]{a^{2} p^{2}+b^{2}}\) Where, \(p=\frac{d y}{d x}\) \(\text { Now, } y=\frac{d y}{d x} x+x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(y-\frac{d y}{d x} x=x\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]^{\frac{1}{3}}\) \(\quad\left[y-\frac{d y}{d x} x\right]^{3}=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(y^{3}-3 y^{2} \frac{d y}{d x}+3 y\left[\frac{d y}{d x}\right]^{2}-x^{3}\left[\frac{d y}{d x}\right]^{3}\) \(=x^{3}\left[a^{2}\left[\frac{d y}{d x}\right]^{2}+b^{2}\right]\) \(\therefore \quad(a-b)^{3}=a^{3}-3 a^{2} b+3 a b^{2}-b^{3}\) Hence, highest order of derivative \(=1\) The power of the highest order derivative \(=3\)
