87292
The solution of the differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) is
1 \(\log ((2 x-4 y)+3)=x-2 y+c\)
2 \(\log [2(2 x-4 y)+3]=2(x-2 y)+c\)
3 \(\log [2(x-2 y)+5]=2(x+y)+c\)
4 \(\log [4(x-2 y)+5]=4(x+2 y)+c\) (c is an arbitrary constant)
Explanation:
(D) : We have differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) \(\frac{d y}{d x}=-\frac{(x-2 y+1)}{(2 x-4 y+3)} \Rightarrow \frac{d y}{d x}=-\frac{(x-2 y+1)}{2(x-2 y)+3}\) Let, \(\quad x-2 y=v\) \(1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{d y}{d x}\) Putting in equation we get- \(\frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{-(v+1)}{2 v+3} \Rightarrow 1-\frac{-2(v+1)}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3+2 v+2}{2 v+3}=\frac{d v}{d x} \Rightarrow \frac{4 v+5}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3}{4 v+5} d v=d x\) Integrating on both side we get- \(\int \frac{2 v+3}{4 v+5} d v=\int d x\) \(\frac{1}{2} \int \frac{4 v+5+1}{4 v+5} d v=\int d x\) \(\frac{1}{2}\left(1+\frac{1}{4 v+5}\right) d v=\int d x\) \(v+\frac{1}{4} \log (4 v+5)=2 x+c\) \(\frac{1}{4} \log (4 v+5)=2 x-v+c\) \(\frac{1}{4} \log [4(x-2 y)+5]=2 x-(x-2 y)+c\) \(\log 4(x-2 y)+5=4(x+2 y)+c\)
AP EAMCET-2016
Differential Equation
87294
The solution of the differential equation \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) is
1 \(\sec ^{3} y^{2}=(\ln x)^{2}+c\)
2 \(\sec ^{3} y^{2}=12(\ln x)+c\)
3 \(\tan y^{2}=(\ln x)^{2}+c\)
4 None of the above where \(\mathrm{c}\) is an arbitrary constant.
Explanation:
(C): \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{4}=\ln x^{2}+\ln \left(e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}=\ln x^{2}+y^{4}-y^{4}\) \(2 y \sec ^{2}\left(y^{2}\right) d y=\frac{\ln x^{2}}{x} d x\) Now, Integrated both side, we get, \(\int 2 y \sec ^{2}\left(y^{2}\right) d y=\frac{1}{2} \int \frac{2 \ln x^{2}}{x} d x\) Let \(\mathrm{y}^{2}=\mathrm{a} \quad \& \quad\) Let \(\ln \mathrm{x}^{2}=\mathrm{b}\) \(2 \mathrm{ydy}=2 \mathrm{da} \quad \frac{1}{\mathrm{x}^{2}} \cdot 2 \mathrm{xdx}=\mathrm{db}\) So, \(\int \sec ^{2}(a) d a=\frac{1}{2} \int b d b\) \(\tan \mathrm{a}=\frac{1}{2} \cdot \frac{\mathrm{b}^{2}}{2}+\mathrm{c} \quad \Rightarrow \tan \mathrm{y}^{2}=\frac{1}{2}\left(\frac{\left(\ln \mathrm{x}^{2}\right)^{2}}{2}\right)+\mathrm{c}\) \(\tan y^{2}=\frac{1}{4}\left(\ln x^{2}\right)^{2}+c \Rightarrow \tan y^{2}=\frac{1}{4}(2 \ln x)^{2}+c\) \(\tan \mathrm{y}^{2} \quad \ln \mathrm{x}^{2} \quad \mathrm{c}\)
SCRA-2014
Differential Equation
87295
The differential equation \(y \frac{d y}{d x}+x=a\) where \(a\) is a constant, represents?
1 a set of circles having centre on the y-axis
2 a set of parabolas
3 a set of circles having centre on the \(x\)-axis
4 a set of straight lines
Explanation:
(C): Given, \(y \frac{d y}{d x}+x=a \Rightarrow \int y d y+\int x d x=\int a d x\) \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=a x+c \Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}-a x=c\) \(\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{ax}=2 \mathrm{c}\) Similar as:- \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) So, the differential equation is a set of circles having centre on the x-axis.
SCRA-2014
Differential Equation
87296
If \(I_{1}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}-y=x^{2}\) and \(I_{2}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}+y=x^{-2}\), then which one of the following is not correct?
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87292
The solution of the differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) is
1 \(\log ((2 x-4 y)+3)=x-2 y+c\)
2 \(\log [2(2 x-4 y)+3]=2(x-2 y)+c\)
3 \(\log [2(x-2 y)+5]=2(x+y)+c\)
4 \(\log [4(x-2 y)+5]=4(x+2 y)+c\) (c is an arbitrary constant)
Explanation:
(D) : We have differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) \(\frac{d y}{d x}=-\frac{(x-2 y+1)}{(2 x-4 y+3)} \Rightarrow \frac{d y}{d x}=-\frac{(x-2 y+1)}{2(x-2 y)+3}\) Let, \(\quad x-2 y=v\) \(1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{d y}{d x}\) Putting in equation we get- \(\frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{-(v+1)}{2 v+3} \Rightarrow 1-\frac{-2(v+1)}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3+2 v+2}{2 v+3}=\frac{d v}{d x} \Rightarrow \frac{4 v+5}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3}{4 v+5} d v=d x\) Integrating on both side we get- \(\int \frac{2 v+3}{4 v+5} d v=\int d x\) \(\frac{1}{2} \int \frac{4 v+5+1}{4 v+5} d v=\int d x\) \(\frac{1}{2}\left(1+\frac{1}{4 v+5}\right) d v=\int d x\) \(v+\frac{1}{4} \log (4 v+5)=2 x+c\) \(\frac{1}{4} \log (4 v+5)=2 x-v+c\) \(\frac{1}{4} \log [4(x-2 y)+5]=2 x-(x-2 y)+c\) \(\log 4(x-2 y)+5=4(x+2 y)+c\)
AP EAMCET-2016
Differential Equation
87294
The solution of the differential equation \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) is
1 \(\sec ^{3} y^{2}=(\ln x)^{2}+c\)
2 \(\sec ^{3} y^{2}=12(\ln x)+c\)
3 \(\tan y^{2}=(\ln x)^{2}+c\)
4 None of the above where \(\mathrm{c}\) is an arbitrary constant.
Explanation:
(C): \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{4}=\ln x^{2}+\ln \left(e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}=\ln x^{2}+y^{4}-y^{4}\) \(2 y \sec ^{2}\left(y^{2}\right) d y=\frac{\ln x^{2}}{x} d x\) Now, Integrated both side, we get, \(\int 2 y \sec ^{2}\left(y^{2}\right) d y=\frac{1}{2} \int \frac{2 \ln x^{2}}{x} d x\) Let \(\mathrm{y}^{2}=\mathrm{a} \quad \& \quad\) Let \(\ln \mathrm{x}^{2}=\mathrm{b}\) \(2 \mathrm{ydy}=2 \mathrm{da} \quad \frac{1}{\mathrm{x}^{2}} \cdot 2 \mathrm{xdx}=\mathrm{db}\) So, \(\int \sec ^{2}(a) d a=\frac{1}{2} \int b d b\) \(\tan \mathrm{a}=\frac{1}{2} \cdot \frac{\mathrm{b}^{2}}{2}+\mathrm{c} \quad \Rightarrow \tan \mathrm{y}^{2}=\frac{1}{2}\left(\frac{\left(\ln \mathrm{x}^{2}\right)^{2}}{2}\right)+\mathrm{c}\) \(\tan y^{2}=\frac{1}{4}\left(\ln x^{2}\right)^{2}+c \Rightarrow \tan y^{2}=\frac{1}{4}(2 \ln x)^{2}+c\) \(\tan \mathrm{y}^{2} \quad \ln \mathrm{x}^{2} \quad \mathrm{c}\)
SCRA-2014
Differential Equation
87295
The differential equation \(y \frac{d y}{d x}+x=a\) where \(a\) is a constant, represents?
1 a set of circles having centre on the y-axis
2 a set of parabolas
3 a set of circles having centre on the \(x\)-axis
4 a set of straight lines
Explanation:
(C): Given, \(y \frac{d y}{d x}+x=a \Rightarrow \int y d y+\int x d x=\int a d x\) \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=a x+c \Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}-a x=c\) \(\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{ax}=2 \mathrm{c}\) Similar as:- \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) So, the differential equation is a set of circles having centre on the x-axis.
SCRA-2014
Differential Equation
87296
If \(I_{1}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}-y=x^{2}\) and \(I_{2}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}+y=x^{-2}\), then which one of the following is not correct?
87292
The solution of the differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) is
1 \(\log ((2 x-4 y)+3)=x-2 y+c\)
2 \(\log [2(2 x-4 y)+3]=2(x-2 y)+c\)
3 \(\log [2(x-2 y)+5]=2(x+y)+c\)
4 \(\log [4(x-2 y)+5]=4(x+2 y)+c\) (c is an arbitrary constant)
Explanation:
(D) : We have differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) \(\frac{d y}{d x}=-\frac{(x-2 y+1)}{(2 x-4 y+3)} \Rightarrow \frac{d y}{d x}=-\frac{(x-2 y+1)}{2(x-2 y)+3}\) Let, \(\quad x-2 y=v\) \(1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{d y}{d x}\) Putting in equation we get- \(\frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{-(v+1)}{2 v+3} \Rightarrow 1-\frac{-2(v+1)}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3+2 v+2}{2 v+3}=\frac{d v}{d x} \Rightarrow \frac{4 v+5}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3}{4 v+5} d v=d x\) Integrating on both side we get- \(\int \frac{2 v+3}{4 v+5} d v=\int d x\) \(\frac{1}{2} \int \frac{4 v+5+1}{4 v+5} d v=\int d x\) \(\frac{1}{2}\left(1+\frac{1}{4 v+5}\right) d v=\int d x\) \(v+\frac{1}{4} \log (4 v+5)=2 x+c\) \(\frac{1}{4} \log (4 v+5)=2 x-v+c\) \(\frac{1}{4} \log [4(x-2 y)+5]=2 x-(x-2 y)+c\) \(\log 4(x-2 y)+5=4(x+2 y)+c\)
AP EAMCET-2016
Differential Equation
87294
The solution of the differential equation \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) is
1 \(\sec ^{3} y^{2}=(\ln x)^{2}+c\)
2 \(\sec ^{3} y^{2}=12(\ln x)+c\)
3 \(\tan y^{2}=(\ln x)^{2}+c\)
4 None of the above where \(\mathrm{c}\) is an arbitrary constant.
Explanation:
(C): \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{4}=\ln x^{2}+\ln \left(e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}=\ln x^{2}+y^{4}-y^{4}\) \(2 y \sec ^{2}\left(y^{2}\right) d y=\frac{\ln x^{2}}{x} d x\) Now, Integrated both side, we get, \(\int 2 y \sec ^{2}\left(y^{2}\right) d y=\frac{1}{2} \int \frac{2 \ln x^{2}}{x} d x\) Let \(\mathrm{y}^{2}=\mathrm{a} \quad \& \quad\) Let \(\ln \mathrm{x}^{2}=\mathrm{b}\) \(2 \mathrm{ydy}=2 \mathrm{da} \quad \frac{1}{\mathrm{x}^{2}} \cdot 2 \mathrm{xdx}=\mathrm{db}\) So, \(\int \sec ^{2}(a) d a=\frac{1}{2} \int b d b\) \(\tan \mathrm{a}=\frac{1}{2} \cdot \frac{\mathrm{b}^{2}}{2}+\mathrm{c} \quad \Rightarrow \tan \mathrm{y}^{2}=\frac{1}{2}\left(\frac{\left(\ln \mathrm{x}^{2}\right)^{2}}{2}\right)+\mathrm{c}\) \(\tan y^{2}=\frac{1}{4}\left(\ln x^{2}\right)^{2}+c \Rightarrow \tan y^{2}=\frac{1}{4}(2 \ln x)^{2}+c\) \(\tan \mathrm{y}^{2} \quad \ln \mathrm{x}^{2} \quad \mathrm{c}\)
SCRA-2014
Differential Equation
87295
The differential equation \(y \frac{d y}{d x}+x=a\) where \(a\) is a constant, represents?
1 a set of circles having centre on the y-axis
2 a set of parabolas
3 a set of circles having centre on the \(x\)-axis
4 a set of straight lines
Explanation:
(C): Given, \(y \frac{d y}{d x}+x=a \Rightarrow \int y d y+\int x d x=\int a d x\) \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=a x+c \Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}-a x=c\) \(\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{ax}=2 \mathrm{c}\) Similar as:- \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) So, the differential equation is a set of circles having centre on the x-axis.
SCRA-2014
Differential Equation
87296
If \(I_{1}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}-y=x^{2}\) and \(I_{2}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}+y=x^{-2}\), then which one of the following is not correct?
87292
The solution of the differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) is
1 \(\log ((2 x-4 y)+3)=x-2 y+c\)
2 \(\log [2(2 x-4 y)+3]=2(x-2 y)+c\)
3 \(\log [2(x-2 y)+5]=2(x+y)+c\)
4 \(\log [4(x-2 y)+5]=4(x+2 y)+c\) (c is an arbitrary constant)
Explanation:
(D) : We have differential equation \((2 x-4 y+3) \frac{d y}{d x}+(x-2 y+1)=0\) \(\frac{d y}{d x}=-\frac{(x-2 y+1)}{(2 x-4 y+3)} \Rightarrow \frac{d y}{d x}=-\frac{(x-2 y+1)}{2(x-2 y)+3}\) Let, \(\quad x-2 y=v\) \(1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{d y}{d x}\) Putting in equation we get- \(\frac{1}{2}\left(1-\frac{d v}{d x}\right)=\frac{-(v+1)}{2 v+3} \Rightarrow 1-\frac{-2(v+1)}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3+2 v+2}{2 v+3}=\frac{d v}{d x} \Rightarrow \frac{4 v+5}{2 v+3}=\frac{d v}{d x}\) \(\frac{2 v+3}{4 v+5} d v=d x\) Integrating on both side we get- \(\int \frac{2 v+3}{4 v+5} d v=\int d x\) \(\frac{1}{2} \int \frac{4 v+5+1}{4 v+5} d v=\int d x\) \(\frac{1}{2}\left(1+\frac{1}{4 v+5}\right) d v=\int d x\) \(v+\frac{1}{4} \log (4 v+5)=2 x+c\) \(\frac{1}{4} \log (4 v+5)=2 x-v+c\) \(\frac{1}{4} \log [4(x-2 y)+5]=2 x-(x-2 y)+c\) \(\log 4(x-2 y)+5=4(x+2 y)+c\)
AP EAMCET-2016
Differential Equation
87294
The solution of the differential equation \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) is
1 \(\sec ^{3} y^{2}=(\ln x)^{2}+c\)
2 \(\sec ^{3} y^{2}=12(\ln x)+c\)
3 \(\tan y^{2}=(\ln x)^{2}+c\)
4 None of the above where \(\mathrm{c}\) is an arbitrary constant.
Explanation:
(C): \(y\left[2 x \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{3}\right]=\ln \left(x^{2} e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}+y^{4}=\ln x^{2}+\ln \left(e^{y^{4}}\right)\) \(2 x y \sec ^{2}\left(y^{2}\right) \frac{d y}{d x}=\ln x^{2}+y^{4}-y^{4}\) \(2 y \sec ^{2}\left(y^{2}\right) d y=\frac{\ln x^{2}}{x} d x\) Now, Integrated both side, we get, \(\int 2 y \sec ^{2}\left(y^{2}\right) d y=\frac{1}{2} \int \frac{2 \ln x^{2}}{x} d x\) Let \(\mathrm{y}^{2}=\mathrm{a} \quad \& \quad\) Let \(\ln \mathrm{x}^{2}=\mathrm{b}\) \(2 \mathrm{ydy}=2 \mathrm{da} \quad \frac{1}{\mathrm{x}^{2}} \cdot 2 \mathrm{xdx}=\mathrm{db}\) So, \(\int \sec ^{2}(a) d a=\frac{1}{2} \int b d b\) \(\tan \mathrm{a}=\frac{1}{2} \cdot \frac{\mathrm{b}^{2}}{2}+\mathrm{c} \quad \Rightarrow \tan \mathrm{y}^{2}=\frac{1}{2}\left(\frac{\left(\ln \mathrm{x}^{2}\right)^{2}}{2}\right)+\mathrm{c}\) \(\tan y^{2}=\frac{1}{4}\left(\ln x^{2}\right)^{2}+c \Rightarrow \tan y^{2}=\frac{1}{4}(2 \ln x)^{2}+c\) \(\tan \mathrm{y}^{2} \quad \ln \mathrm{x}^{2} \quad \mathrm{c}\)
SCRA-2014
Differential Equation
87295
The differential equation \(y \frac{d y}{d x}+x=a\) where \(a\) is a constant, represents?
1 a set of circles having centre on the y-axis
2 a set of parabolas
3 a set of circles having centre on the \(x\)-axis
4 a set of straight lines
Explanation:
(C): Given, \(y \frac{d y}{d x}+x=a \Rightarrow \int y d y+\int x d x=\int a d x\) \(\frac{y^{2}}{2}+\frac{x^{2}}{2}=a x+c \Rightarrow \frac{y^{2}}{2}+\frac{x^{2}}{2}-a x=c\) \(\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{ax}=2 \mathrm{c}\) Similar as:- \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) So, the differential equation is a set of circles having centre on the x-axis.
SCRA-2014
Differential Equation
87296
If \(I_{1}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}-y=x^{2}\) and \(I_{2}\) is the integrating factor of the differential equation \(x \frac{d y}{d x}+y=x^{-2}\), then which one of the following is not correct?