87621
If \(y=y(x)\) is the solution curve of the differential equation \(\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1\) then \(y\left(\frac{\pi}{6}\right)\) is equal to
87623
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^{2}(e)\) is \(=\)
1 \(\mathrm{e}^{2}\)
2 \(\frac{3}{2} \mathrm{e}^{2}\)
3 \(3 \mathrm{e}^{2}\)
4 \(2 \mathrm{e}^{2}\)
Explanation:
(D) : Given, differential equation - \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\) \(\frac{3 y^{2}-x^{2}}{3 x y}=\frac{d y}{d x} \tag{i}\) Which is a homogeneous type, Let's consider \(\mathrm{y}=\mathrm{vx}\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i) \(\frac{3(v x)^{2}-x^{2}}{3 x(v x)}=v+x \frac{d v}{d x}\) \(\frac{3 v^{2}-1}{3 v}-v=\frac{x d v}{d x}\) \(\frac{-1}{3 v}=\frac{x d v}{d x}\) \(\int \frac{d x}{x}=-\int 3 v d v\) On integrating both sides, \(\log x=-3 \frac{v^{2}}{2}+c\) Which satisfies \(y(1) \stackrel{2}{=} 1\) \(\therefore \quad \log (1)=\frac{-3}{2}\left(\frac{1}{1}\right)^{2}+\mathrm{c}\) \(\mathrm{c}=\frac{3}{2}\) \(\therefore\) Equation, \(\log x=\frac{-3}{2}\left(\frac{y}{x}\right)^{2}+\frac{3}{2}\) \(3 y^{2}=2 x^{2}\left(\frac{3}{2}-\log x\right)\) \(6 y^{2}=6 x^{2}-4 x^{2} \log x\) \(\therefore \quad\) For \(\mathrm{x}=\mathrm{e}\) \(\therefore \quad 6 \mathrm{y}^{2}(\mathrm{e})=6 \mathrm{e}_{2}^{2}-4 \mathrm{e}^{2} \cdot \log \mathrm{e}\)
JEE Main-24.01.2023
Differential Equation
87624
Let \(y=y(x)\) be the solution of the differential equation \(\log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) If \(y(2)=2\), then \(y(e)\) is equal to
1 \(\frac{4+\mathrm{e}^{2}}{4}\)
2 \(\frac{1+\mathrm{e}^{2}}{2}\)
3 \(\frac{2+\mathrm{e}^{2}}{2}\)
4 \(\frac{1+\mathrm{e}^{2}}{4}\)
Explanation:
(A) : Given, differential equation- \(x \log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{y}{x \log x}=x\) IF \(=e^{\int \frac{d x}{x \log x}}=e^{\log |\log x|}=|\log x|\) Solution of equation is, \(y(\text { IF })=\int x \cdot|\log x| d x\) \(y|\log x|=|\log x| \frac{x^{2}}{2}-\frac{x^{2}}{4}+c\) On putting \(\mathrm{x}=2\) \(2 \log 2=\log 2.2-1+c\) Put, \(\mathrm{x}=\mathrm{e}\) \(y=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1\) \(y(e)=1+\frac{e^{2}}{4}\) \(y(e)=\frac{4+e^{2}}{4}\)
JEE Main-29.01.2023
Differential Equation
87625
If \(x=x(y)\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to :
87621
If \(y=y(x)\) is the solution curve of the differential equation \(\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1\) then \(y\left(\frac{\pi}{6}\right)\) is equal to
87623
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^{2}(e)\) is \(=\)
1 \(\mathrm{e}^{2}\)
2 \(\frac{3}{2} \mathrm{e}^{2}\)
3 \(3 \mathrm{e}^{2}\)
4 \(2 \mathrm{e}^{2}\)
Explanation:
(D) : Given, differential equation - \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\) \(\frac{3 y^{2}-x^{2}}{3 x y}=\frac{d y}{d x} \tag{i}\) Which is a homogeneous type, Let's consider \(\mathrm{y}=\mathrm{vx}\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i) \(\frac{3(v x)^{2}-x^{2}}{3 x(v x)}=v+x \frac{d v}{d x}\) \(\frac{3 v^{2}-1}{3 v}-v=\frac{x d v}{d x}\) \(\frac{-1}{3 v}=\frac{x d v}{d x}\) \(\int \frac{d x}{x}=-\int 3 v d v\) On integrating both sides, \(\log x=-3 \frac{v^{2}}{2}+c\) Which satisfies \(y(1) \stackrel{2}{=} 1\) \(\therefore \quad \log (1)=\frac{-3}{2}\left(\frac{1}{1}\right)^{2}+\mathrm{c}\) \(\mathrm{c}=\frac{3}{2}\) \(\therefore\) Equation, \(\log x=\frac{-3}{2}\left(\frac{y}{x}\right)^{2}+\frac{3}{2}\) \(3 y^{2}=2 x^{2}\left(\frac{3}{2}-\log x\right)\) \(6 y^{2}=6 x^{2}-4 x^{2} \log x\) \(\therefore \quad\) For \(\mathrm{x}=\mathrm{e}\) \(\therefore \quad 6 \mathrm{y}^{2}(\mathrm{e})=6 \mathrm{e}_{2}^{2}-4 \mathrm{e}^{2} \cdot \log \mathrm{e}\)
JEE Main-24.01.2023
Differential Equation
87624
Let \(y=y(x)\) be the solution of the differential equation \(\log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) If \(y(2)=2\), then \(y(e)\) is equal to
1 \(\frac{4+\mathrm{e}^{2}}{4}\)
2 \(\frac{1+\mathrm{e}^{2}}{2}\)
3 \(\frac{2+\mathrm{e}^{2}}{2}\)
4 \(\frac{1+\mathrm{e}^{2}}{4}\)
Explanation:
(A) : Given, differential equation- \(x \log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{y}{x \log x}=x\) IF \(=e^{\int \frac{d x}{x \log x}}=e^{\log |\log x|}=|\log x|\) Solution of equation is, \(y(\text { IF })=\int x \cdot|\log x| d x\) \(y|\log x|=|\log x| \frac{x^{2}}{2}-\frac{x^{2}}{4}+c\) On putting \(\mathrm{x}=2\) \(2 \log 2=\log 2.2-1+c\) Put, \(\mathrm{x}=\mathrm{e}\) \(y=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1\) \(y(e)=1+\frac{e^{2}}{4}\) \(y(e)=\frac{4+e^{2}}{4}\)
JEE Main-29.01.2023
Differential Equation
87625
If \(x=x(y)\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to :
87621
If \(y=y(x)\) is the solution curve of the differential equation \(\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1\) then \(y\left(\frac{\pi}{6}\right)\) is equal to
87623
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^{2}(e)\) is \(=\)
1 \(\mathrm{e}^{2}\)
2 \(\frac{3}{2} \mathrm{e}^{2}\)
3 \(3 \mathrm{e}^{2}\)
4 \(2 \mathrm{e}^{2}\)
Explanation:
(D) : Given, differential equation - \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\) \(\frac{3 y^{2}-x^{2}}{3 x y}=\frac{d y}{d x} \tag{i}\) Which is a homogeneous type, Let's consider \(\mathrm{y}=\mathrm{vx}\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i) \(\frac{3(v x)^{2}-x^{2}}{3 x(v x)}=v+x \frac{d v}{d x}\) \(\frac{3 v^{2}-1}{3 v}-v=\frac{x d v}{d x}\) \(\frac{-1}{3 v}=\frac{x d v}{d x}\) \(\int \frac{d x}{x}=-\int 3 v d v\) On integrating both sides, \(\log x=-3 \frac{v^{2}}{2}+c\) Which satisfies \(y(1) \stackrel{2}{=} 1\) \(\therefore \quad \log (1)=\frac{-3}{2}\left(\frac{1}{1}\right)^{2}+\mathrm{c}\) \(\mathrm{c}=\frac{3}{2}\) \(\therefore\) Equation, \(\log x=\frac{-3}{2}\left(\frac{y}{x}\right)^{2}+\frac{3}{2}\) \(3 y^{2}=2 x^{2}\left(\frac{3}{2}-\log x\right)\) \(6 y^{2}=6 x^{2}-4 x^{2} \log x\) \(\therefore \quad\) For \(\mathrm{x}=\mathrm{e}\) \(\therefore \quad 6 \mathrm{y}^{2}(\mathrm{e})=6 \mathrm{e}_{2}^{2}-4 \mathrm{e}^{2} \cdot \log \mathrm{e}\)
JEE Main-24.01.2023
Differential Equation
87624
Let \(y=y(x)\) be the solution of the differential equation \(\log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) If \(y(2)=2\), then \(y(e)\) is equal to
1 \(\frac{4+\mathrm{e}^{2}}{4}\)
2 \(\frac{1+\mathrm{e}^{2}}{2}\)
3 \(\frac{2+\mathrm{e}^{2}}{2}\)
4 \(\frac{1+\mathrm{e}^{2}}{4}\)
Explanation:
(A) : Given, differential equation- \(x \log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{y}{x \log x}=x\) IF \(=e^{\int \frac{d x}{x \log x}}=e^{\log |\log x|}=|\log x|\) Solution of equation is, \(y(\text { IF })=\int x \cdot|\log x| d x\) \(y|\log x|=|\log x| \frac{x^{2}}{2}-\frac{x^{2}}{4}+c\) On putting \(\mathrm{x}=2\) \(2 \log 2=\log 2.2-1+c\) Put, \(\mathrm{x}=\mathrm{e}\) \(y=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1\) \(y(e)=1+\frac{e^{2}}{4}\) \(y(e)=\frac{4+e^{2}}{4}\)
JEE Main-29.01.2023
Differential Equation
87625
If \(x=x(y)\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to :
87621
If \(y=y(x)\) is the solution curve of the differential equation \(\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1\) then \(y\left(\frac{\pi}{6}\right)\) is equal to
87623
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^{2}(e)\) is \(=\)
1 \(\mathrm{e}^{2}\)
2 \(\frac{3}{2} \mathrm{e}^{2}\)
3 \(3 \mathrm{e}^{2}\)
4 \(2 \mathrm{e}^{2}\)
Explanation:
(D) : Given, differential equation - \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\) \(\frac{3 y^{2}-x^{2}}{3 x y}=\frac{d y}{d x} \tag{i}\) Which is a homogeneous type, Let's consider \(\mathrm{y}=\mathrm{vx}\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i) \(\frac{3(v x)^{2}-x^{2}}{3 x(v x)}=v+x \frac{d v}{d x}\) \(\frac{3 v^{2}-1}{3 v}-v=\frac{x d v}{d x}\) \(\frac{-1}{3 v}=\frac{x d v}{d x}\) \(\int \frac{d x}{x}=-\int 3 v d v\) On integrating both sides, \(\log x=-3 \frac{v^{2}}{2}+c\) Which satisfies \(y(1) \stackrel{2}{=} 1\) \(\therefore \quad \log (1)=\frac{-3}{2}\left(\frac{1}{1}\right)^{2}+\mathrm{c}\) \(\mathrm{c}=\frac{3}{2}\) \(\therefore\) Equation, \(\log x=\frac{-3}{2}\left(\frac{y}{x}\right)^{2}+\frac{3}{2}\) \(3 y^{2}=2 x^{2}\left(\frac{3}{2}-\log x\right)\) \(6 y^{2}=6 x^{2}-4 x^{2} \log x\) \(\therefore \quad\) For \(\mathrm{x}=\mathrm{e}\) \(\therefore \quad 6 \mathrm{y}^{2}(\mathrm{e})=6 \mathrm{e}_{2}^{2}-4 \mathrm{e}^{2} \cdot \log \mathrm{e}\)
JEE Main-24.01.2023
Differential Equation
87624
Let \(y=y(x)\) be the solution of the differential equation \(\log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) If \(y(2)=2\), then \(y(e)\) is equal to
1 \(\frac{4+\mathrm{e}^{2}}{4}\)
2 \(\frac{1+\mathrm{e}^{2}}{2}\)
3 \(\frac{2+\mathrm{e}^{2}}{2}\)
4 \(\frac{1+\mathrm{e}^{2}}{4}\)
Explanation:
(A) : Given, differential equation- \(x \log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{y}{x \log x}=x\) IF \(=e^{\int \frac{d x}{x \log x}}=e^{\log |\log x|}=|\log x|\) Solution of equation is, \(y(\text { IF })=\int x \cdot|\log x| d x\) \(y|\log x|=|\log x| \frac{x^{2}}{2}-\frac{x^{2}}{4}+c\) On putting \(\mathrm{x}=2\) \(2 \log 2=\log 2.2-1+c\) Put, \(\mathrm{x}=\mathrm{e}\) \(y=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1\) \(y(e)=1+\frac{e^{2}}{4}\) \(y(e)=\frac{4+e^{2}}{4}\)
JEE Main-29.01.2023
Differential Equation
87625
If \(x=x(y)\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to :
87621
If \(y=y(x)\) is the solution curve of the differential equation \(\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1\) then \(y\left(\frac{\pi}{6}\right)\) is equal to
87623
Let \(y=y(x)\) be the solution of the differential equation \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\). Then \(6 y^{2}(e)\) is \(=\)
1 \(\mathrm{e}^{2}\)
2 \(\frac{3}{2} \mathrm{e}^{2}\)
3 \(3 \mathrm{e}^{2}\)
4 \(2 \mathrm{e}^{2}\)
Explanation:
(D) : Given, differential equation - \(\left(x^{2}-3 y^{2}\right) d x+3 x y d y=0, y(1)=1\) \(\frac{3 y^{2}-x^{2}}{3 x y}=\frac{d y}{d x} \tag{i}\) Which is a homogeneous type, Let's consider \(\mathrm{y}=\mathrm{vx}\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}\) From equation (i) \(\frac{3(v x)^{2}-x^{2}}{3 x(v x)}=v+x \frac{d v}{d x}\) \(\frac{3 v^{2}-1}{3 v}-v=\frac{x d v}{d x}\) \(\frac{-1}{3 v}=\frac{x d v}{d x}\) \(\int \frac{d x}{x}=-\int 3 v d v\) On integrating both sides, \(\log x=-3 \frac{v^{2}}{2}+c\) Which satisfies \(y(1) \stackrel{2}{=} 1\) \(\therefore \quad \log (1)=\frac{-3}{2}\left(\frac{1}{1}\right)^{2}+\mathrm{c}\) \(\mathrm{c}=\frac{3}{2}\) \(\therefore\) Equation, \(\log x=\frac{-3}{2}\left(\frac{y}{x}\right)^{2}+\frac{3}{2}\) \(3 y^{2}=2 x^{2}\left(\frac{3}{2}-\log x\right)\) \(6 y^{2}=6 x^{2}-4 x^{2} \log x\) \(\therefore \quad\) For \(\mathrm{x}=\mathrm{e}\) \(\therefore \quad 6 \mathrm{y}^{2}(\mathrm{e})=6 \mathrm{e}_{2}^{2}-4 \mathrm{e}^{2} \cdot \log \mathrm{e}\)
JEE Main-24.01.2023
Differential Equation
87624
Let \(y=y(x)\) be the solution of the differential equation \(\log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) If \(y(2)=2\), then \(y(e)\) is equal to
1 \(\frac{4+\mathrm{e}^{2}}{4}\)
2 \(\frac{1+\mathrm{e}^{2}}{2}\)
3 \(\frac{2+\mathrm{e}^{2}}{2}\)
4 \(\frac{1+\mathrm{e}^{2}}{4}\)
Explanation:
(A) : Given, differential equation- \(x \log _{e} x \frac{d y}{d x}+y=x^{2} \log _{e} x(x>1)\) \(\frac{d y}{d x}+\frac{y}{x \log x}=x\) IF \(=e^{\int \frac{d x}{x \log x}}=e^{\log |\log x|}=|\log x|\) Solution of equation is, \(y(\text { IF })=\int x \cdot|\log x| d x\) \(y|\log x|=|\log x| \frac{x^{2}}{2}-\frac{x^{2}}{4}+c\) On putting \(\mathrm{x}=2\) \(2 \log 2=\log 2.2-1+c\) Put, \(\mathrm{x}=\mathrm{e}\) \(y=\frac{e^{2}}{2}-\frac{e^{2}}{4}+1\) \(y(e)=1+\frac{e^{2}}{4}\) \(y(e)=\frac{4+e^{2}}{4}\)
JEE Main-29.01.2023
Differential Equation
87625
If \(x=x(y)\) is the solution of the differential equation \(y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0\); then \(x(e)\) is equal to :
