(A) : Given, Let, \(\frac{x+y-1}{x+y-2} \cdot \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\) \(\mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\left(\frac{d v}{d x}-1\right)\) Then, \(\quad \frac{v-1}{v-2}\left(\frac{d v}{d x}-1\right)=\frac{v+1}{v+2}\) \(\frac{d v}{d x}=\left(\frac{v+1}{v+2}\right)\left(\frac{v-2}{v-1}\right)+1 \Rightarrow \frac{d v}{d x}=\frac{2 v^{2}-4}{v^{2}+v-2}\) \(\frac{v^{2}+v-2}{2 v^{2}-4} d v=d x \Rightarrow \int \frac{v^{2}+v-2}{v^{2}-2} d v=\int 2 d x\) \(\int\left(1+\frac{v}{v^{2}-2}\right) d v=\int 2 d x\) \(v+\frac{1}{2} \log \left(v^{2}-2\right)=2 x+c\) Now, putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\mathrm{x}+\mathrm{y}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=2 \mathrm{x}+\mathrm{c}\) \(\mathrm{y}-\mathrm{x}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=\mathrm{c} \tag{i}\) Given, \(y=1 \text { when } x=1\) \(1-1+\frac{1}{2} \log |4-2|=c \Rightarrow \frac{1}{2} \log 2=c\) On putting the value \(\mathrm{c}\) in equation (i), \(y-x+\frac{1}{2} \log \left|(x+y)^{2}-2\right|=\frac{1}{2} \log 2\) \(y-x+\frac{1}{2} \log \left|\frac{(x+y)^{2}-2}{2}\right|=0\) \(2(y-x)+\log \left|\frac{(x+y)^{2}-2}{2}\right|=0\)
AP EAMCET-22.04.2019
Differential Equation
87316
If \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) where \(y(0)=1, y^{\prime}(0)=0\), then the value of \(y\) at \(x=\log _{e} 2\) is
1 1
2 -1
3 2
4 0
Explanation:
(D) : Given, \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) \(\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0 \tag{i}\) Let, \(\mathrm{m}=\mathrm{dy} / \mathrm{dx}\) Then, \(\quad m^{2}-3 m+2=0\) \((\mathrm{m}-1)(\mathrm{m}-2)=0\) \(\mathrm{m}=1,2\) General solution of differential equation \(y=A e^{x}+B e^{2 x}\) \(y^{\prime}=A e^{x}+2 B^{2 x}\) Given, \(y(0)=1\) \(1=A+B \tag{ii}\) And, \(\quad y^{\prime}(0)=0\) or \(x=0\) then \(y=0\) \(\mathrm{A}=2 \text { and } \mathrm{B}=-1\) Substituting the value of \(A\) and \(B\) then \(y=2 e^{x}-e^{2 x}\) At, \(x=\log _{e} 2\) \(y=2 e^{\log _{e} 2}-e^{2 \log _{e} 2}\) \(y=2 \times 2-2^{2}=0\)
WB JEE-2010
Differential Equation
87317
Let \(y\) be the solution of the differential equation \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}\) satisfying \(y(1)=1\). Then, \(y\) satisfies
1 \(y=x^{y-1}\)
2 \(y=x^{y}\)
3 \(y=x^{y+1}\)
4 \(y=x^{y+2}\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x} \Rightarrow \frac{d x}{d y}=\frac{x(1-y \log x)}{y^{2}}\) \(\frac{d x}{d y}-\frac{x}{y^{2}}=\frac{-x y \log x}{y^{2}} \Rightarrow \frac{1}{x} \frac{d x}{d y}-\frac{1}{y^{2}}=-\frac{\log x}{y}\) Let, \(\log x=t\) \(\frac{1}{x} \frac{d x}{d y}=\frac{d t}{d y}\) Now, \(\frac{\mathrm{dt}}{\mathrm{dy}}-\frac{1}{\mathrm{y}^{2}}=-\frac{\mathrm{t}}{\mathrm{y}} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dy}}+\frac{\mathrm{t}}{\mathrm{y}}=\frac{1}{\mathrm{y}^{2}}\) Now it is linear differential equation I.F, \(\quad=\mathrm{e}^{\int \frac{1}{y} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}\) I.F, \(\quad=\mathrm{y}\) For general solution- \(\text { t.y }=\int \frac{1}{y^{2}} \cdot y d y \Rightarrow t \cdot y=\int \frac{1}{y} \cdot d y\) \(\text { t. } y=\log y+C\) Putting \(\mathrm{t}=\log \mathrm{x}\) Given, \(y \log x=\log y+C\) Then, \(y(1)=1\) \(\mathrm{C}=0\) \(y \log x=\log y\) \(\log x^{y}=\log y\) \(y=x^{y}\)
WB JEE-2012
Differential Equation
87334
If \(y=e^{-x} \cos 2 x\), then which of the following differential equation is satisfied?
(A) : Given, Let, \(\frac{x+y-1}{x+y-2} \cdot \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\) \(\mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\left(\frac{d v}{d x}-1\right)\) Then, \(\quad \frac{v-1}{v-2}\left(\frac{d v}{d x}-1\right)=\frac{v+1}{v+2}\) \(\frac{d v}{d x}=\left(\frac{v+1}{v+2}\right)\left(\frac{v-2}{v-1}\right)+1 \Rightarrow \frac{d v}{d x}=\frac{2 v^{2}-4}{v^{2}+v-2}\) \(\frac{v^{2}+v-2}{2 v^{2}-4} d v=d x \Rightarrow \int \frac{v^{2}+v-2}{v^{2}-2} d v=\int 2 d x\) \(\int\left(1+\frac{v}{v^{2}-2}\right) d v=\int 2 d x\) \(v+\frac{1}{2} \log \left(v^{2}-2\right)=2 x+c\) Now, putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\mathrm{x}+\mathrm{y}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=2 \mathrm{x}+\mathrm{c}\) \(\mathrm{y}-\mathrm{x}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=\mathrm{c} \tag{i}\) Given, \(y=1 \text { when } x=1\) \(1-1+\frac{1}{2} \log |4-2|=c \Rightarrow \frac{1}{2} \log 2=c\) On putting the value \(\mathrm{c}\) in equation (i), \(y-x+\frac{1}{2} \log \left|(x+y)^{2}-2\right|=\frac{1}{2} \log 2\) \(y-x+\frac{1}{2} \log \left|\frac{(x+y)^{2}-2}{2}\right|=0\) \(2(y-x)+\log \left|\frac{(x+y)^{2}-2}{2}\right|=0\)
AP EAMCET-22.04.2019
Differential Equation
87316
If \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) where \(y(0)=1, y^{\prime}(0)=0\), then the value of \(y\) at \(x=\log _{e} 2\) is
1 1
2 -1
3 2
4 0
Explanation:
(D) : Given, \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) \(\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0 \tag{i}\) Let, \(\mathrm{m}=\mathrm{dy} / \mathrm{dx}\) Then, \(\quad m^{2}-3 m+2=0\) \((\mathrm{m}-1)(\mathrm{m}-2)=0\) \(\mathrm{m}=1,2\) General solution of differential equation \(y=A e^{x}+B e^{2 x}\) \(y^{\prime}=A e^{x}+2 B^{2 x}\) Given, \(y(0)=1\) \(1=A+B \tag{ii}\) And, \(\quad y^{\prime}(0)=0\) or \(x=0\) then \(y=0\) \(\mathrm{A}=2 \text { and } \mathrm{B}=-1\) Substituting the value of \(A\) and \(B\) then \(y=2 e^{x}-e^{2 x}\) At, \(x=\log _{e} 2\) \(y=2 e^{\log _{e} 2}-e^{2 \log _{e} 2}\) \(y=2 \times 2-2^{2}=0\)
WB JEE-2010
Differential Equation
87317
Let \(y\) be the solution of the differential equation \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}\) satisfying \(y(1)=1\). Then, \(y\) satisfies
1 \(y=x^{y-1}\)
2 \(y=x^{y}\)
3 \(y=x^{y+1}\)
4 \(y=x^{y+2}\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x} \Rightarrow \frac{d x}{d y}=\frac{x(1-y \log x)}{y^{2}}\) \(\frac{d x}{d y}-\frac{x}{y^{2}}=\frac{-x y \log x}{y^{2}} \Rightarrow \frac{1}{x} \frac{d x}{d y}-\frac{1}{y^{2}}=-\frac{\log x}{y}\) Let, \(\log x=t\) \(\frac{1}{x} \frac{d x}{d y}=\frac{d t}{d y}\) Now, \(\frac{\mathrm{dt}}{\mathrm{dy}}-\frac{1}{\mathrm{y}^{2}}=-\frac{\mathrm{t}}{\mathrm{y}} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dy}}+\frac{\mathrm{t}}{\mathrm{y}}=\frac{1}{\mathrm{y}^{2}}\) Now it is linear differential equation I.F, \(\quad=\mathrm{e}^{\int \frac{1}{y} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}\) I.F, \(\quad=\mathrm{y}\) For general solution- \(\text { t.y }=\int \frac{1}{y^{2}} \cdot y d y \Rightarrow t \cdot y=\int \frac{1}{y} \cdot d y\) \(\text { t. } y=\log y+C\) Putting \(\mathrm{t}=\log \mathrm{x}\) Given, \(y \log x=\log y+C\) Then, \(y(1)=1\) \(\mathrm{C}=0\) \(y \log x=\log y\) \(\log x^{y}=\log y\) \(y=x^{y}\)
WB JEE-2012
Differential Equation
87334
If \(y=e^{-x} \cos 2 x\), then which of the following differential equation is satisfied?
(A) : Given, Let, \(\frac{x+y-1}{x+y-2} \cdot \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\) \(\mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\left(\frac{d v}{d x}-1\right)\) Then, \(\quad \frac{v-1}{v-2}\left(\frac{d v}{d x}-1\right)=\frac{v+1}{v+2}\) \(\frac{d v}{d x}=\left(\frac{v+1}{v+2}\right)\left(\frac{v-2}{v-1}\right)+1 \Rightarrow \frac{d v}{d x}=\frac{2 v^{2}-4}{v^{2}+v-2}\) \(\frac{v^{2}+v-2}{2 v^{2}-4} d v=d x \Rightarrow \int \frac{v^{2}+v-2}{v^{2}-2} d v=\int 2 d x\) \(\int\left(1+\frac{v}{v^{2}-2}\right) d v=\int 2 d x\) \(v+\frac{1}{2} \log \left(v^{2}-2\right)=2 x+c\) Now, putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\mathrm{x}+\mathrm{y}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=2 \mathrm{x}+\mathrm{c}\) \(\mathrm{y}-\mathrm{x}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=\mathrm{c} \tag{i}\) Given, \(y=1 \text { when } x=1\) \(1-1+\frac{1}{2} \log |4-2|=c \Rightarrow \frac{1}{2} \log 2=c\) On putting the value \(\mathrm{c}\) in equation (i), \(y-x+\frac{1}{2} \log \left|(x+y)^{2}-2\right|=\frac{1}{2} \log 2\) \(y-x+\frac{1}{2} \log \left|\frac{(x+y)^{2}-2}{2}\right|=0\) \(2(y-x)+\log \left|\frac{(x+y)^{2}-2}{2}\right|=0\)
AP EAMCET-22.04.2019
Differential Equation
87316
If \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) where \(y(0)=1, y^{\prime}(0)=0\), then the value of \(y\) at \(x=\log _{e} 2\) is
1 1
2 -1
3 2
4 0
Explanation:
(D) : Given, \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) \(\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0 \tag{i}\) Let, \(\mathrm{m}=\mathrm{dy} / \mathrm{dx}\) Then, \(\quad m^{2}-3 m+2=0\) \((\mathrm{m}-1)(\mathrm{m}-2)=0\) \(\mathrm{m}=1,2\) General solution of differential equation \(y=A e^{x}+B e^{2 x}\) \(y^{\prime}=A e^{x}+2 B^{2 x}\) Given, \(y(0)=1\) \(1=A+B \tag{ii}\) And, \(\quad y^{\prime}(0)=0\) or \(x=0\) then \(y=0\) \(\mathrm{A}=2 \text { and } \mathrm{B}=-1\) Substituting the value of \(A\) and \(B\) then \(y=2 e^{x}-e^{2 x}\) At, \(x=\log _{e} 2\) \(y=2 e^{\log _{e} 2}-e^{2 \log _{e} 2}\) \(y=2 \times 2-2^{2}=0\)
WB JEE-2010
Differential Equation
87317
Let \(y\) be the solution of the differential equation \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}\) satisfying \(y(1)=1\). Then, \(y\) satisfies
1 \(y=x^{y-1}\)
2 \(y=x^{y}\)
3 \(y=x^{y+1}\)
4 \(y=x^{y+2}\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x} \Rightarrow \frac{d x}{d y}=\frac{x(1-y \log x)}{y^{2}}\) \(\frac{d x}{d y}-\frac{x}{y^{2}}=\frac{-x y \log x}{y^{2}} \Rightarrow \frac{1}{x} \frac{d x}{d y}-\frac{1}{y^{2}}=-\frac{\log x}{y}\) Let, \(\log x=t\) \(\frac{1}{x} \frac{d x}{d y}=\frac{d t}{d y}\) Now, \(\frac{\mathrm{dt}}{\mathrm{dy}}-\frac{1}{\mathrm{y}^{2}}=-\frac{\mathrm{t}}{\mathrm{y}} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dy}}+\frac{\mathrm{t}}{\mathrm{y}}=\frac{1}{\mathrm{y}^{2}}\) Now it is linear differential equation I.F, \(\quad=\mathrm{e}^{\int \frac{1}{y} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}\) I.F, \(\quad=\mathrm{y}\) For general solution- \(\text { t.y }=\int \frac{1}{y^{2}} \cdot y d y \Rightarrow t \cdot y=\int \frac{1}{y} \cdot d y\) \(\text { t. } y=\log y+C\) Putting \(\mathrm{t}=\log \mathrm{x}\) Given, \(y \log x=\log y+C\) Then, \(y(1)=1\) \(\mathrm{C}=0\) \(y \log x=\log y\) \(\log x^{y}=\log y\) \(y=x^{y}\)
WB JEE-2012
Differential Equation
87334
If \(y=e^{-x} \cos 2 x\), then which of the following differential equation is satisfied?
(A) : Given, Let, \(\frac{x+y-1}{x+y-2} \cdot \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\) \(\mathrm{x}+\mathrm{y}=\mathrm{v}\) \(1+\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\left(\frac{d v}{d x}-1\right)\) Then, \(\quad \frac{v-1}{v-2}\left(\frac{d v}{d x}-1\right)=\frac{v+1}{v+2}\) \(\frac{d v}{d x}=\left(\frac{v+1}{v+2}\right)\left(\frac{v-2}{v-1}\right)+1 \Rightarrow \frac{d v}{d x}=\frac{2 v^{2}-4}{v^{2}+v-2}\) \(\frac{v^{2}+v-2}{2 v^{2}-4} d v=d x \Rightarrow \int \frac{v^{2}+v-2}{v^{2}-2} d v=\int 2 d x\) \(\int\left(1+\frac{v}{v^{2}-2}\right) d v=\int 2 d x\) \(v+\frac{1}{2} \log \left(v^{2}-2\right)=2 x+c\) Now, putting \(\mathrm{v}=\mathrm{x}+\mathrm{y}\) we get - \(\mathrm{x}+\mathrm{y}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=2 \mathrm{x}+\mathrm{c}\) \(\mathrm{y}-\mathrm{x}+\frac{1}{2} \log \left|(\mathrm{x}+\mathrm{y})^{2}-2\right|=\mathrm{c} \tag{i}\) Given, \(y=1 \text { when } x=1\) \(1-1+\frac{1}{2} \log |4-2|=c \Rightarrow \frac{1}{2} \log 2=c\) On putting the value \(\mathrm{c}\) in equation (i), \(y-x+\frac{1}{2} \log \left|(x+y)^{2}-2\right|=\frac{1}{2} \log 2\) \(y-x+\frac{1}{2} \log \left|\frac{(x+y)^{2}-2}{2}\right|=0\) \(2(y-x)+\log \left|\frac{(x+y)^{2}-2}{2}\right|=0\)
AP EAMCET-22.04.2019
Differential Equation
87316
If \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) where \(y(0)=1, y^{\prime}(0)=0\), then the value of \(y\) at \(x=\log _{e} 2\) is
1 1
2 -1
3 2
4 0
Explanation:
(D) : Given, \(y^{\prime \prime}-3 y^{\prime}+2 y=0\) \(\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0 \tag{i}\) Let, \(\mathrm{m}=\mathrm{dy} / \mathrm{dx}\) Then, \(\quad m^{2}-3 m+2=0\) \((\mathrm{m}-1)(\mathrm{m}-2)=0\) \(\mathrm{m}=1,2\) General solution of differential equation \(y=A e^{x}+B e^{2 x}\) \(y^{\prime}=A e^{x}+2 B^{2 x}\) Given, \(y(0)=1\) \(1=A+B \tag{ii}\) And, \(\quad y^{\prime}(0)=0\) or \(x=0\) then \(y=0\) \(\mathrm{A}=2 \text { and } \mathrm{B}=-1\) Substituting the value of \(A\) and \(B\) then \(y=2 e^{x}-e^{2 x}\) At, \(x=\log _{e} 2\) \(y=2 e^{\log _{e} 2}-e^{2 \log _{e} 2}\) \(y=2 \times 2-2^{2}=0\)
WB JEE-2010
Differential Equation
87317
Let \(y\) be the solution of the differential equation \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}\) satisfying \(y(1)=1\). Then, \(y\) satisfies
1 \(y=x^{y-1}\)
2 \(y=x^{y}\)
3 \(y=x^{y+1}\)
4 \(y=x^{y+2}\)
Explanation:
(B) : Given, \(x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x} \Rightarrow \frac{d x}{d y}=\frac{x(1-y \log x)}{y^{2}}\) \(\frac{d x}{d y}-\frac{x}{y^{2}}=\frac{-x y \log x}{y^{2}} \Rightarrow \frac{1}{x} \frac{d x}{d y}-\frac{1}{y^{2}}=-\frac{\log x}{y}\) Let, \(\log x=t\) \(\frac{1}{x} \frac{d x}{d y}=\frac{d t}{d y}\) Now, \(\frac{\mathrm{dt}}{\mathrm{dy}}-\frac{1}{\mathrm{y}^{2}}=-\frac{\mathrm{t}}{\mathrm{y}} \Rightarrow \frac{\mathrm{dt}}{\mathrm{dy}}+\frac{\mathrm{t}}{\mathrm{y}}=\frac{1}{\mathrm{y}^{2}}\) Now it is linear differential equation I.F, \(\quad=\mathrm{e}^{\int \frac{1}{y} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}\) I.F, \(\quad=\mathrm{y}\) For general solution- \(\text { t.y }=\int \frac{1}{y^{2}} \cdot y d y \Rightarrow t \cdot y=\int \frac{1}{y} \cdot d y\) \(\text { t. } y=\log y+C\) Putting \(\mathrm{t}=\log \mathrm{x}\) Given, \(y \log x=\log y+C\) Then, \(y(1)=1\) \(\mathrm{C}=0\) \(y \log x=\log y\) \(\log x^{y}=\log y\) \(y=x^{y}\)
WB JEE-2012
Differential Equation
87334
If \(y=e^{-x} \cos 2 x\), then which of the following differential equation is satisfied?
