NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87189
The differential equation of all parabolas whose axes are parallel to the axis of \(y\), is
1 \(\frac{d^{3} y}{d x^{3}}=1\)
2 \(\frac{d^{3} y}{d x^{3}}=-1\)
3 \(\frac{d^{3} y}{d x^{3}}=0\)
4 none of these
Explanation:
(C) : The equation of a member of the family of parabolas having axis parallel to \(\mathrm{y}\)-axis is \(y=A x^{2}+B x+c\) Where, A, B and C are arbitrary constant. Differentiating both side w.r.t. \(x\), we get- \(\frac{d y}{d x}=2 A x+B \Rightarrow \frac{d^{2} y}{d x^{2}}=2 A\) \(\frac{d^{3} y}{d x^{3}}=0\)
SRM JEEE-2008
Differential Equation
87168
If \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0 y(1)=1\), then \(y(2)\) is equal to :
1 \(2+\log _{2} 3\)
2 \(2+\log _{2} 2\)
3 \(2-\log _{2} 2\)
4 \(2-\log _{2} 3\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0, y(1)=1\) \(\frac{d y}{d x}=\frac{-2^{x}\left(2^{y}-1\right)}{2^{y}\left(2^{x}-1\right)} \Rightarrow \int \frac{2^{y}}{2^{y}-1} d y=-\int \frac{2^{x}}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \int \frac{2^{y} \ln 2}{2^{y}-1} d y=-\frac{1}{\ln 2} \int \frac{2^{x} \ln 2}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \ln \left|2^{y}-1\right|=-\frac{1}{\ln 2} \ln \left|2^{x}-1\right|+C\) At \(\quad \mathrm{x}=1, \mathrm{y}=1\) Put given value, we get \(\mathrm{c}=0\) \(\ln \left|2^{y}-1\right|+\ln \left|2^{x}-1\right|=0\) \(2^{y}-1=\frac{1}{2^{x}+1}\) At \(\quad x=1 \Rightarrow 2^{y}=\frac{1}{3}+1 \Rightarrow 2^{y}=\frac{4}{3}\) So, \(\quad y=\log _{2} \frac{4}{3} \Rightarrow \mathrm{y}=\log _{2} 4-\log _{2} 3\) \(y=2-\log _{2} 3\)
JEE Main-27.06.2022
Differential Equation
87169
The general solution of the differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) is
1 \(y=x^{2} \sin x+c\)
2 \(y=\sin x+c x^{2}\)
3 \(y=\cos x+c x^{2}\)
4 \(y=x^{2} \sin x+c x^{2}\)
Explanation:
(D) : Given differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) \(x^{2} d y=2 x y d x+x^{4} \cos x d x\) \(x^{2} d y=\left(2 x y+x^{4} \cos x\right) d x\) \(\frac{d y}{d x}=\frac{x^{4} \cos x+2 x y}{x^{2}} \Rightarrow \frac{d y}{d x}=x^{2} \cos x+\frac{2 y}{x}\) \(\frac{d y}{d x}-\frac{2 y}{x}=x^{2} \cos x\) It is form of \(\frac{d y}{d x}+P y=Q\) \(P=\frac{-2}{x} \text { and } Q=x^{2} \cos x \Rightarrow\) \(I . F=e^{\int P d x} \Rightarrow=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^{2}}\) Now general solution - \(\text { y.I.F }=\int \text { Q.I.F }+c\) \(y \cdot\left(\frac{1}{x^{2}}\right)=\int\left(x^{2} \cos x\right) \frac{1}{x^{2}} d x+c\) \(\frac{y}{x^{2}}=\sin x+c\) \(y=x^{2} \sin x+c x^{2}\)
Karnataka CET-2020
Differential Equation
87170
The solution for the differential equation \(\frac{d y}{y}+\frac{d x}{x}=0\) is
(C) : Given that, differential equation \(\frac{\mathrm{dy}}{\mathrm{y}}+\frac{\mathrm{dx}}{\mathrm{x}}=0\) Now, integrating both sides, we get - \(\log y+\log x=\log c \Rightarrow \log x y=\log c\) \(x y=c\)
87189
The differential equation of all parabolas whose axes are parallel to the axis of \(y\), is
1 \(\frac{d^{3} y}{d x^{3}}=1\)
2 \(\frac{d^{3} y}{d x^{3}}=-1\)
3 \(\frac{d^{3} y}{d x^{3}}=0\)
4 none of these
Explanation:
(C) : The equation of a member of the family of parabolas having axis parallel to \(\mathrm{y}\)-axis is \(y=A x^{2}+B x+c\) Where, A, B and C are arbitrary constant. Differentiating both side w.r.t. \(x\), we get- \(\frac{d y}{d x}=2 A x+B \Rightarrow \frac{d^{2} y}{d x^{2}}=2 A\) \(\frac{d^{3} y}{d x^{3}}=0\)
SRM JEEE-2008
Differential Equation
87168
If \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0 y(1)=1\), then \(y(2)\) is equal to :
1 \(2+\log _{2} 3\)
2 \(2+\log _{2} 2\)
3 \(2-\log _{2} 2\)
4 \(2-\log _{2} 3\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0, y(1)=1\) \(\frac{d y}{d x}=\frac{-2^{x}\left(2^{y}-1\right)}{2^{y}\left(2^{x}-1\right)} \Rightarrow \int \frac{2^{y}}{2^{y}-1} d y=-\int \frac{2^{x}}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \int \frac{2^{y} \ln 2}{2^{y}-1} d y=-\frac{1}{\ln 2} \int \frac{2^{x} \ln 2}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \ln \left|2^{y}-1\right|=-\frac{1}{\ln 2} \ln \left|2^{x}-1\right|+C\) At \(\quad \mathrm{x}=1, \mathrm{y}=1\) Put given value, we get \(\mathrm{c}=0\) \(\ln \left|2^{y}-1\right|+\ln \left|2^{x}-1\right|=0\) \(2^{y}-1=\frac{1}{2^{x}+1}\) At \(\quad x=1 \Rightarrow 2^{y}=\frac{1}{3}+1 \Rightarrow 2^{y}=\frac{4}{3}\) So, \(\quad y=\log _{2} \frac{4}{3} \Rightarrow \mathrm{y}=\log _{2} 4-\log _{2} 3\) \(y=2-\log _{2} 3\)
JEE Main-27.06.2022
Differential Equation
87169
The general solution of the differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) is
1 \(y=x^{2} \sin x+c\)
2 \(y=\sin x+c x^{2}\)
3 \(y=\cos x+c x^{2}\)
4 \(y=x^{2} \sin x+c x^{2}\)
Explanation:
(D) : Given differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) \(x^{2} d y=2 x y d x+x^{4} \cos x d x\) \(x^{2} d y=\left(2 x y+x^{4} \cos x\right) d x\) \(\frac{d y}{d x}=\frac{x^{4} \cos x+2 x y}{x^{2}} \Rightarrow \frac{d y}{d x}=x^{2} \cos x+\frac{2 y}{x}\) \(\frac{d y}{d x}-\frac{2 y}{x}=x^{2} \cos x\) It is form of \(\frac{d y}{d x}+P y=Q\) \(P=\frac{-2}{x} \text { and } Q=x^{2} \cos x \Rightarrow\) \(I . F=e^{\int P d x} \Rightarrow=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^{2}}\) Now general solution - \(\text { y.I.F }=\int \text { Q.I.F }+c\) \(y \cdot\left(\frac{1}{x^{2}}\right)=\int\left(x^{2} \cos x\right) \frac{1}{x^{2}} d x+c\) \(\frac{y}{x^{2}}=\sin x+c\) \(y=x^{2} \sin x+c x^{2}\)
Karnataka CET-2020
Differential Equation
87170
The solution for the differential equation \(\frac{d y}{y}+\frac{d x}{x}=0\) is
(C) : Given that, differential equation \(\frac{\mathrm{dy}}{\mathrm{y}}+\frac{\mathrm{dx}}{\mathrm{x}}=0\) Now, integrating both sides, we get - \(\log y+\log x=\log c \Rightarrow \log x y=\log c\) \(x y=c\)
87189
The differential equation of all parabolas whose axes are parallel to the axis of \(y\), is
1 \(\frac{d^{3} y}{d x^{3}}=1\)
2 \(\frac{d^{3} y}{d x^{3}}=-1\)
3 \(\frac{d^{3} y}{d x^{3}}=0\)
4 none of these
Explanation:
(C) : The equation of a member of the family of parabolas having axis parallel to \(\mathrm{y}\)-axis is \(y=A x^{2}+B x+c\) Where, A, B and C are arbitrary constant. Differentiating both side w.r.t. \(x\), we get- \(\frac{d y}{d x}=2 A x+B \Rightarrow \frac{d^{2} y}{d x^{2}}=2 A\) \(\frac{d^{3} y}{d x^{3}}=0\)
SRM JEEE-2008
Differential Equation
87168
If \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0 y(1)=1\), then \(y(2)\) is equal to :
1 \(2+\log _{2} 3\)
2 \(2+\log _{2} 2\)
3 \(2-\log _{2} 2\)
4 \(2-\log _{2} 3\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0, y(1)=1\) \(\frac{d y}{d x}=\frac{-2^{x}\left(2^{y}-1\right)}{2^{y}\left(2^{x}-1\right)} \Rightarrow \int \frac{2^{y}}{2^{y}-1} d y=-\int \frac{2^{x}}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \int \frac{2^{y} \ln 2}{2^{y}-1} d y=-\frac{1}{\ln 2} \int \frac{2^{x} \ln 2}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \ln \left|2^{y}-1\right|=-\frac{1}{\ln 2} \ln \left|2^{x}-1\right|+C\) At \(\quad \mathrm{x}=1, \mathrm{y}=1\) Put given value, we get \(\mathrm{c}=0\) \(\ln \left|2^{y}-1\right|+\ln \left|2^{x}-1\right|=0\) \(2^{y}-1=\frac{1}{2^{x}+1}\) At \(\quad x=1 \Rightarrow 2^{y}=\frac{1}{3}+1 \Rightarrow 2^{y}=\frac{4}{3}\) So, \(\quad y=\log _{2} \frac{4}{3} \Rightarrow \mathrm{y}=\log _{2} 4-\log _{2} 3\) \(y=2-\log _{2} 3\)
JEE Main-27.06.2022
Differential Equation
87169
The general solution of the differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) is
1 \(y=x^{2} \sin x+c\)
2 \(y=\sin x+c x^{2}\)
3 \(y=\cos x+c x^{2}\)
4 \(y=x^{2} \sin x+c x^{2}\)
Explanation:
(D) : Given differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) \(x^{2} d y=2 x y d x+x^{4} \cos x d x\) \(x^{2} d y=\left(2 x y+x^{4} \cos x\right) d x\) \(\frac{d y}{d x}=\frac{x^{4} \cos x+2 x y}{x^{2}} \Rightarrow \frac{d y}{d x}=x^{2} \cos x+\frac{2 y}{x}\) \(\frac{d y}{d x}-\frac{2 y}{x}=x^{2} \cos x\) It is form of \(\frac{d y}{d x}+P y=Q\) \(P=\frac{-2}{x} \text { and } Q=x^{2} \cos x \Rightarrow\) \(I . F=e^{\int P d x} \Rightarrow=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^{2}}\) Now general solution - \(\text { y.I.F }=\int \text { Q.I.F }+c\) \(y \cdot\left(\frac{1}{x^{2}}\right)=\int\left(x^{2} \cos x\right) \frac{1}{x^{2}} d x+c\) \(\frac{y}{x^{2}}=\sin x+c\) \(y=x^{2} \sin x+c x^{2}\)
Karnataka CET-2020
Differential Equation
87170
The solution for the differential equation \(\frac{d y}{y}+\frac{d x}{x}=0\) is
(C) : Given that, differential equation \(\frac{\mathrm{dy}}{\mathrm{y}}+\frac{\mathrm{dx}}{\mathrm{x}}=0\) Now, integrating both sides, we get - \(\log y+\log x=\log c \Rightarrow \log x y=\log c\) \(x y=c\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87189
The differential equation of all parabolas whose axes are parallel to the axis of \(y\), is
1 \(\frac{d^{3} y}{d x^{3}}=1\)
2 \(\frac{d^{3} y}{d x^{3}}=-1\)
3 \(\frac{d^{3} y}{d x^{3}}=0\)
4 none of these
Explanation:
(C) : The equation of a member of the family of parabolas having axis parallel to \(\mathrm{y}\)-axis is \(y=A x^{2}+B x+c\) Where, A, B and C are arbitrary constant. Differentiating both side w.r.t. \(x\), we get- \(\frac{d y}{d x}=2 A x+B \Rightarrow \frac{d^{2} y}{d x^{2}}=2 A\) \(\frac{d^{3} y}{d x^{3}}=0\)
SRM JEEE-2008
Differential Equation
87168
If \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0 y(1)=1\), then \(y(2)\) is equal to :
1 \(2+\log _{2} 3\)
2 \(2+\log _{2} 2\)
3 \(2-\log _{2} 2\)
4 \(2-\log _{2} 3\)
Explanation:
(D) : Given, \(\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y>0, y(1)=1\) \(\frac{d y}{d x}=\frac{-2^{x}\left(2^{y}-1\right)}{2^{y}\left(2^{x}-1\right)} \Rightarrow \int \frac{2^{y}}{2^{y}-1} d y=-\int \frac{2^{x}}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \int \frac{2^{y} \ln 2}{2^{y}-1} d y=-\frac{1}{\ln 2} \int \frac{2^{x} \ln 2}{2^{x}-1} d x\) \(\frac{1}{\ln 2} \ln \left|2^{y}-1\right|=-\frac{1}{\ln 2} \ln \left|2^{x}-1\right|+C\) At \(\quad \mathrm{x}=1, \mathrm{y}=1\) Put given value, we get \(\mathrm{c}=0\) \(\ln \left|2^{y}-1\right|+\ln \left|2^{x}-1\right|=0\) \(2^{y}-1=\frac{1}{2^{x}+1}\) At \(\quad x=1 \Rightarrow 2^{y}=\frac{1}{3}+1 \Rightarrow 2^{y}=\frac{4}{3}\) So, \(\quad y=\log _{2} \frac{4}{3} \Rightarrow \mathrm{y}=\log _{2} 4-\log _{2} 3\) \(y=2-\log _{2} 3\)
JEE Main-27.06.2022
Differential Equation
87169
The general solution of the differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) is
1 \(y=x^{2} \sin x+c\)
2 \(y=\sin x+c x^{2}\)
3 \(y=\cos x+c x^{2}\)
4 \(y=x^{2} \sin x+c x^{2}\)
Explanation:
(D) : Given differential equation \(x^{2} d y-2 x y d x=x^{4} \cos x d x\) \(x^{2} d y=2 x y d x+x^{4} \cos x d x\) \(x^{2} d y=\left(2 x y+x^{4} \cos x\right) d x\) \(\frac{d y}{d x}=\frac{x^{4} \cos x+2 x y}{x^{2}} \Rightarrow \frac{d y}{d x}=x^{2} \cos x+\frac{2 y}{x}\) \(\frac{d y}{d x}-\frac{2 y}{x}=x^{2} \cos x\) It is form of \(\frac{d y}{d x}+P y=Q\) \(P=\frac{-2}{x} \text { and } Q=x^{2} \cos x \Rightarrow\) \(I . F=e^{\int P d x} \Rightarrow=e^{\int \frac{-2}{x} d x}=e^{-2 \log x}=\frac{1}{x^{2}}\) Now general solution - \(\text { y.I.F }=\int \text { Q.I.F }+c\) \(y \cdot\left(\frac{1}{x^{2}}\right)=\int\left(x^{2} \cos x\right) \frac{1}{x^{2}} d x+c\) \(\frac{y}{x^{2}}=\sin x+c\) \(y=x^{2} \sin x+c x^{2}\)
Karnataka CET-2020
Differential Equation
87170
The solution for the differential equation \(\frac{d y}{y}+\frac{d x}{x}=0\) is
(C) : Given that, differential equation \(\frac{\mathrm{dy}}{\mathrm{y}}+\frac{\mathrm{dx}}{\mathrm{x}}=0\) Now, integrating both sides, we get - \(\log y+\log x=\log c \Rightarrow \log x y=\log c\) \(x y=c\)