87477
Let \(y=y(x)\) be the solution of the differential equation, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{4}{9 \sqrt{3}} \pi^{2}\)
2 \(\pi^{2}\)
3 \(-\frac{8}{9} \pi^{2}\)
4 \(-\frac{4}{9} \pi^{2}\)
Explanation:
(C) : Given, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\) \(\therefore \quad \frac{d y}{d x}+y \frac{\cos x}{\sin x}=\frac{4 x}{\sin x}\) \(\frac{d y}{d x}+\cot x \cdot y=\frac{4 x}{\sin x}\) Equation (i) comparing with \(\frac{d y}{d x}+P(x) \cdot y=Q(x)\) \(\therefore \quad \mathrm{P}(\mathrm{x})=\cot \mathrm{x}\) and \(\mathrm{Q}(\mathrm{x})=\frac{4 \mathrm{x}}{\sin \mathrm{x}}\) I.F \(\quad=e^{\int P(x) d x}\) \(=\mathrm{e}^{\int \cot x \mathrm{dx}}\) \(=\mathrm{e}^{\int \log (|\sin x|) d x=\sin x}\) \(\therefore\) Solution is given by \(y \cdot \sin x=\int \frac{4 x}{\sin x} \cdot \sin x d x+c\) \(=\int 4 \mathrm{x} d \mathrm{x}+\mathrm{c}\) \(\begin{aligned} =\frac{4 \mathrm{x}^{2}}{2}+c \\ =2 \mathrm{x}^{2}+\mathrm{c}\end{aligned}\) \(\therefore \quad \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^{2}+\mathrm{c}\) \(\because\) Given \(y\left(\frac{\pi}{2}\right)=0 \quad\), when \(x=\frac{\pi}{2}, y=0\) \(\therefore\) Equation \(0=2 x\left(\frac{\pi}{2}\right)^{2}+\mathrm{c}\) \(\therefore \quad \mathrm{c}=\frac{\pi^{2}}{2}\) Put the value \(\mathrm{c}=\frac{\pi^{2}}{2}\) in equation (ii) \(y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\) Find, \(y\left(\frac{\pi}{6}\right)\) \(\therefore \quad y \sin \frac{\pi}{6}=2\left(\frac{\pi}{6}\right)^{2}-\frac{\pi^{2}}{2}\) \(\mathrm{y} \times \frac{1}{2}=2 \cdot \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2}\) \(\frac{y}{2}=\frac{\pi^{2}}{18}-\frac{\pi^{2}}{2}\) \(y=\frac{-8 \pi^{2}}{9}\) So, \(\quad y\left(\frac{\pi}{6}\right)=-\frac{8}{9} \pi^{2}\)
JEE Main-2018
Differential Equation
87478
Let \(y=y(x)\) be the solution of the differential equation, \(\quad\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is
(D) : Given, \(\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\tan ^{-1} \mathrm{x}-\mathrm{t}}{1+\mathrm{x}^{2}}\) \(\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}}=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}}\) Which is form of \(\frac{d x}{d y}+P x=Q\) I.F \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} \quad\left(\because \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}\right)\) I.F. \(=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
MHT CET-2020
Differential Equation
87479
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad\) such that \(y(0)=1\). Then
(A) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytan} x=2 \mathrm{x}+\mathrm{x}^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) Linear differential equation in the form of \(\frac{d y}{d x}+P y=Q\) \(\therefore I F=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}}(\sec \mathrm{x})}=\sec \mathrm{x}\) Now, solution of linear differential equation is given as, \(y \times I F=\int(Q \times I F) d x+C\) \(\therefore \quad y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C\) \(=\int(2 x \operatorname{sex}) d x+\int x^{2} \sec x \tan x d x+C\) \(\because \quad \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x\) Therefore, \(y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int_{0} x \sec x d x+C\) \(y \sec x=x^{2} \sec x+C \tag{i}\) \(\because \quad \mathrm{y}(0)=1 \Rightarrow 1(1)=0(1)+C\) \(\therefore \mathrm{C}=1\) \(\ddot{\text { Now, }} \mathrm{y}=\mathrm{x}^{2}+\cos \mathrm{x}\) from equation (i) and \(y^{\prime}=2 x-\sin x\) Now, \(\mathrm{y}^{\prime}\left(\frac{\pi}{4}\right)-\mathrm{y}^{\prime}\left(-\frac{\pi}{4}\right)=\left[2\left(\frac{\pi}{4}\right)-\frac{1}{\sqrt{2}}\right]-\left[2\left(-\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\right]=\pi-\sqrt{2}\)
Differential Equation
87480
If \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{\pi^{2}}{2 \sqrt{3}}\)
2 \(-\frac{\pi^{2}}{2 \sqrt{3}}\)
3 \(-\frac{\pi^{2}}{4 \sqrt{3}}\)
4 \(-\frac{\pi^{2}}{2}\)
Explanation:
(B) : Given, \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\) \(\Rightarrow \quad \frac{d y}{d x}-y \frac{\sin x}{\cos x}=\frac{6 x}{\cos x}\) \(\Rightarrow \quad \frac{d y}{d x}-y \tan x=\frac{6 x}{\cos x}\) Which is the linear differential equation of the form \(\frac{d y}{d x}+P x=Q\) So, \(\mathrm{IF}=\mathrm{e}^{-\int \tan x \mathrm{dx}}=\mathrm{e}^{-\log (\sec x)}=\cos \mathrm{x}\) \(\therefore \quad\) Required solution of differential equation is \(y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+c\) \(\because\) Given that, \(y\left(\frac{\pi}{3}\right)=0\) So, \(\quad 0=3\left(\frac{\pi}{3}\right)^{2}+\mathrm{C}\) \(\mathrm{C}=-\frac{\pi^{2}}{3}\) \(\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}\) Now, at \(\mathrm{x}=\frac{\pi}{6}\) \(y\left(\frac{\sqrt{3}}{2}\right)=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4}\) Hence, \(y=-\frac{\pi^{2}}{2 \sqrt{3}}\)
87477
Let \(y=y(x)\) be the solution of the differential equation, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{4}{9 \sqrt{3}} \pi^{2}\)
2 \(\pi^{2}\)
3 \(-\frac{8}{9} \pi^{2}\)
4 \(-\frac{4}{9} \pi^{2}\)
Explanation:
(C) : Given, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\) \(\therefore \quad \frac{d y}{d x}+y \frac{\cos x}{\sin x}=\frac{4 x}{\sin x}\) \(\frac{d y}{d x}+\cot x \cdot y=\frac{4 x}{\sin x}\) Equation (i) comparing with \(\frac{d y}{d x}+P(x) \cdot y=Q(x)\) \(\therefore \quad \mathrm{P}(\mathrm{x})=\cot \mathrm{x}\) and \(\mathrm{Q}(\mathrm{x})=\frac{4 \mathrm{x}}{\sin \mathrm{x}}\) I.F \(\quad=e^{\int P(x) d x}\) \(=\mathrm{e}^{\int \cot x \mathrm{dx}}\) \(=\mathrm{e}^{\int \log (|\sin x|) d x=\sin x}\) \(\therefore\) Solution is given by \(y \cdot \sin x=\int \frac{4 x}{\sin x} \cdot \sin x d x+c\) \(=\int 4 \mathrm{x} d \mathrm{x}+\mathrm{c}\) \(\begin{aligned} =\frac{4 \mathrm{x}^{2}}{2}+c \\ =2 \mathrm{x}^{2}+\mathrm{c}\end{aligned}\) \(\therefore \quad \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^{2}+\mathrm{c}\) \(\because\) Given \(y\left(\frac{\pi}{2}\right)=0 \quad\), when \(x=\frac{\pi}{2}, y=0\) \(\therefore\) Equation \(0=2 x\left(\frac{\pi}{2}\right)^{2}+\mathrm{c}\) \(\therefore \quad \mathrm{c}=\frac{\pi^{2}}{2}\) Put the value \(\mathrm{c}=\frac{\pi^{2}}{2}\) in equation (ii) \(y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\) Find, \(y\left(\frac{\pi}{6}\right)\) \(\therefore \quad y \sin \frac{\pi}{6}=2\left(\frac{\pi}{6}\right)^{2}-\frac{\pi^{2}}{2}\) \(\mathrm{y} \times \frac{1}{2}=2 \cdot \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2}\) \(\frac{y}{2}=\frac{\pi^{2}}{18}-\frac{\pi^{2}}{2}\) \(y=\frac{-8 \pi^{2}}{9}\) So, \(\quad y\left(\frac{\pi}{6}\right)=-\frac{8}{9} \pi^{2}\)
JEE Main-2018
Differential Equation
87478
Let \(y=y(x)\) be the solution of the differential equation, \(\quad\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is
(D) : Given, \(\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\tan ^{-1} \mathrm{x}-\mathrm{t}}{1+\mathrm{x}^{2}}\) \(\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}}=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}}\) Which is form of \(\frac{d x}{d y}+P x=Q\) I.F \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} \quad\left(\because \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}\right)\) I.F. \(=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
MHT CET-2020
Differential Equation
87479
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad\) such that \(y(0)=1\). Then
(A) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytan} x=2 \mathrm{x}+\mathrm{x}^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) Linear differential equation in the form of \(\frac{d y}{d x}+P y=Q\) \(\therefore I F=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}}(\sec \mathrm{x})}=\sec \mathrm{x}\) Now, solution of linear differential equation is given as, \(y \times I F=\int(Q \times I F) d x+C\) \(\therefore \quad y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C\) \(=\int(2 x \operatorname{sex}) d x+\int x^{2} \sec x \tan x d x+C\) \(\because \quad \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x\) Therefore, \(y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int_{0} x \sec x d x+C\) \(y \sec x=x^{2} \sec x+C \tag{i}\) \(\because \quad \mathrm{y}(0)=1 \Rightarrow 1(1)=0(1)+C\) \(\therefore \mathrm{C}=1\) \(\ddot{\text { Now, }} \mathrm{y}=\mathrm{x}^{2}+\cos \mathrm{x}\) from equation (i) and \(y^{\prime}=2 x-\sin x\) Now, \(\mathrm{y}^{\prime}\left(\frac{\pi}{4}\right)-\mathrm{y}^{\prime}\left(-\frac{\pi}{4}\right)=\left[2\left(\frac{\pi}{4}\right)-\frac{1}{\sqrt{2}}\right]-\left[2\left(-\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\right]=\pi-\sqrt{2}\)
Differential Equation
87480
If \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{\pi^{2}}{2 \sqrt{3}}\)
2 \(-\frac{\pi^{2}}{2 \sqrt{3}}\)
3 \(-\frac{\pi^{2}}{4 \sqrt{3}}\)
4 \(-\frac{\pi^{2}}{2}\)
Explanation:
(B) : Given, \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\) \(\Rightarrow \quad \frac{d y}{d x}-y \frac{\sin x}{\cos x}=\frac{6 x}{\cos x}\) \(\Rightarrow \quad \frac{d y}{d x}-y \tan x=\frac{6 x}{\cos x}\) Which is the linear differential equation of the form \(\frac{d y}{d x}+P x=Q\) So, \(\mathrm{IF}=\mathrm{e}^{-\int \tan x \mathrm{dx}}=\mathrm{e}^{-\log (\sec x)}=\cos \mathrm{x}\) \(\therefore \quad\) Required solution of differential equation is \(y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+c\) \(\because\) Given that, \(y\left(\frac{\pi}{3}\right)=0\) So, \(\quad 0=3\left(\frac{\pi}{3}\right)^{2}+\mathrm{C}\) \(\mathrm{C}=-\frac{\pi^{2}}{3}\) \(\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}\) Now, at \(\mathrm{x}=\frac{\pi}{6}\) \(y\left(\frac{\sqrt{3}}{2}\right)=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4}\) Hence, \(y=-\frac{\pi^{2}}{2 \sqrt{3}}\)
87477
Let \(y=y(x)\) be the solution of the differential equation, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{4}{9 \sqrt{3}} \pi^{2}\)
2 \(\pi^{2}\)
3 \(-\frac{8}{9} \pi^{2}\)
4 \(-\frac{4}{9} \pi^{2}\)
Explanation:
(C) : Given, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\) \(\therefore \quad \frac{d y}{d x}+y \frac{\cos x}{\sin x}=\frac{4 x}{\sin x}\) \(\frac{d y}{d x}+\cot x \cdot y=\frac{4 x}{\sin x}\) Equation (i) comparing with \(\frac{d y}{d x}+P(x) \cdot y=Q(x)\) \(\therefore \quad \mathrm{P}(\mathrm{x})=\cot \mathrm{x}\) and \(\mathrm{Q}(\mathrm{x})=\frac{4 \mathrm{x}}{\sin \mathrm{x}}\) I.F \(\quad=e^{\int P(x) d x}\) \(=\mathrm{e}^{\int \cot x \mathrm{dx}}\) \(=\mathrm{e}^{\int \log (|\sin x|) d x=\sin x}\) \(\therefore\) Solution is given by \(y \cdot \sin x=\int \frac{4 x}{\sin x} \cdot \sin x d x+c\) \(=\int 4 \mathrm{x} d \mathrm{x}+\mathrm{c}\) \(\begin{aligned} =\frac{4 \mathrm{x}^{2}}{2}+c \\ =2 \mathrm{x}^{2}+\mathrm{c}\end{aligned}\) \(\therefore \quad \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^{2}+\mathrm{c}\) \(\because\) Given \(y\left(\frac{\pi}{2}\right)=0 \quad\), when \(x=\frac{\pi}{2}, y=0\) \(\therefore\) Equation \(0=2 x\left(\frac{\pi}{2}\right)^{2}+\mathrm{c}\) \(\therefore \quad \mathrm{c}=\frac{\pi^{2}}{2}\) Put the value \(\mathrm{c}=\frac{\pi^{2}}{2}\) in equation (ii) \(y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\) Find, \(y\left(\frac{\pi}{6}\right)\) \(\therefore \quad y \sin \frac{\pi}{6}=2\left(\frac{\pi}{6}\right)^{2}-\frac{\pi^{2}}{2}\) \(\mathrm{y} \times \frac{1}{2}=2 \cdot \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2}\) \(\frac{y}{2}=\frac{\pi^{2}}{18}-\frac{\pi^{2}}{2}\) \(y=\frac{-8 \pi^{2}}{9}\) So, \(\quad y\left(\frac{\pi}{6}\right)=-\frac{8}{9} \pi^{2}\)
JEE Main-2018
Differential Equation
87478
Let \(y=y(x)\) be the solution of the differential equation, \(\quad\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is
(D) : Given, \(\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\tan ^{-1} \mathrm{x}-\mathrm{t}}{1+\mathrm{x}^{2}}\) \(\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}}=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}}\) Which is form of \(\frac{d x}{d y}+P x=Q\) I.F \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} \quad\left(\because \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}\right)\) I.F. \(=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
MHT CET-2020
Differential Equation
87479
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad\) such that \(y(0)=1\). Then
(A) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytan} x=2 \mathrm{x}+\mathrm{x}^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) Linear differential equation in the form of \(\frac{d y}{d x}+P y=Q\) \(\therefore I F=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}}(\sec \mathrm{x})}=\sec \mathrm{x}\) Now, solution of linear differential equation is given as, \(y \times I F=\int(Q \times I F) d x+C\) \(\therefore \quad y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C\) \(=\int(2 x \operatorname{sex}) d x+\int x^{2} \sec x \tan x d x+C\) \(\because \quad \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x\) Therefore, \(y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int_{0} x \sec x d x+C\) \(y \sec x=x^{2} \sec x+C \tag{i}\) \(\because \quad \mathrm{y}(0)=1 \Rightarrow 1(1)=0(1)+C\) \(\therefore \mathrm{C}=1\) \(\ddot{\text { Now, }} \mathrm{y}=\mathrm{x}^{2}+\cos \mathrm{x}\) from equation (i) and \(y^{\prime}=2 x-\sin x\) Now, \(\mathrm{y}^{\prime}\left(\frac{\pi}{4}\right)-\mathrm{y}^{\prime}\left(-\frac{\pi}{4}\right)=\left[2\left(\frac{\pi}{4}\right)-\frac{1}{\sqrt{2}}\right]-\left[2\left(-\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\right]=\pi-\sqrt{2}\)
Differential Equation
87480
If \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{\pi^{2}}{2 \sqrt{3}}\)
2 \(-\frac{\pi^{2}}{2 \sqrt{3}}\)
3 \(-\frac{\pi^{2}}{4 \sqrt{3}}\)
4 \(-\frac{\pi^{2}}{2}\)
Explanation:
(B) : Given, \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\) \(\Rightarrow \quad \frac{d y}{d x}-y \frac{\sin x}{\cos x}=\frac{6 x}{\cos x}\) \(\Rightarrow \quad \frac{d y}{d x}-y \tan x=\frac{6 x}{\cos x}\) Which is the linear differential equation of the form \(\frac{d y}{d x}+P x=Q\) So, \(\mathrm{IF}=\mathrm{e}^{-\int \tan x \mathrm{dx}}=\mathrm{e}^{-\log (\sec x)}=\cos \mathrm{x}\) \(\therefore \quad\) Required solution of differential equation is \(y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+c\) \(\because\) Given that, \(y\left(\frac{\pi}{3}\right)=0\) So, \(\quad 0=3\left(\frac{\pi}{3}\right)^{2}+\mathrm{C}\) \(\mathrm{C}=-\frac{\pi^{2}}{3}\) \(\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}\) Now, at \(\mathrm{x}=\frac{\pi}{6}\) \(y\left(\frac{\sqrt{3}}{2}\right)=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4}\) Hence, \(y=-\frac{\pi^{2}}{2 \sqrt{3}}\)
87477
Let \(y=y(x)\) be the solution of the differential equation, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{4}{9 \sqrt{3}} \pi^{2}\)
2 \(\pi^{2}\)
3 \(-\frac{8}{9} \pi^{2}\)
4 \(-\frac{4}{9} \pi^{2}\)
Explanation:
(C) : Given, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\) \(\therefore \quad \frac{d y}{d x}+y \frac{\cos x}{\sin x}=\frac{4 x}{\sin x}\) \(\frac{d y}{d x}+\cot x \cdot y=\frac{4 x}{\sin x}\) Equation (i) comparing with \(\frac{d y}{d x}+P(x) \cdot y=Q(x)\) \(\therefore \quad \mathrm{P}(\mathrm{x})=\cot \mathrm{x}\) and \(\mathrm{Q}(\mathrm{x})=\frac{4 \mathrm{x}}{\sin \mathrm{x}}\) I.F \(\quad=e^{\int P(x) d x}\) \(=\mathrm{e}^{\int \cot x \mathrm{dx}}\) \(=\mathrm{e}^{\int \log (|\sin x|) d x=\sin x}\) \(\therefore\) Solution is given by \(y \cdot \sin x=\int \frac{4 x}{\sin x} \cdot \sin x d x+c\) \(=\int 4 \mathrm{x} d \mathrm{x}+\mathrm{c}\) \(\begin{aligned} =\frac{4 \mathrm{x}^{2}}{2}+c \\ =2 \mathrm{x}^{2}+\mathrm{c}\end{aligned}\) \(\therefore \quad \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^{2}+\mathrm{c}\) \(\because\) Given \(y\left(\frac{\pi}{2}\right)=0 \quad\), when \(x=\frac{\pi}{2}, y=0\) \(\therefore\) Equation \(0=2 x\left(\frac{\pi}{2}\right)^{2}+\mathrm{c}\) \(\therefore \quad \mathrm{c}=\frac{\pi^{2}}{2}\) Put the value \(\mathrm{c}=\frac{\pi^{2}}{2}\) in equation (ii) \(y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\) Find, \(y\left(\frac{\pi}{6}\right)\) \(\therefore \quad y \sin \frac{\pi}{6}=2\left(\frac{\pi}{6}\right)^{2}-\frac{\pi^{2}}{2}\) \(\mathrm{y} \times \frac{1}{2}=2 \cdot \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2}\) \(\frac{y}{2}=\frac{\pi^{2}}{18}-\frac{\pi^{2}}{2}\) \(y=\frac{-8 \pi^{2}}{9}\) So, \(\quad y\left(\frac{\pi}{6}\right)=-\frac{8}{9} \pi^{2}\)
JEE Main-2018
Differential Equation
87478
Let \(y=y(x)\) be the solution of the differential equation, \(\quad\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is
(D) : Given, \(\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\tan ^{-1} \mathrm{x}-\mathrm{t}}{1+\mathrm{x}^{2}}\) \(\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}}=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}}\) Which is form of \(\frac{d x}{d y}+P x=Q\) I.F \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} \quad\left(\because \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}\right)\) I.F. \(=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
MHT CET-2020
Differential Equation
87479
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad\) such that \(y(0)=1\). Then
(A) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytan} x=2 \mathrm{x}+\mathrm{x}^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) Linear differential equation in the form of \(\frac{d y}{d x}+P y=Q\) \(\therefore I F=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}}(\sec \mathrm{x})}=\sec \mathrm{x}\) Now, solution of linear differential equation is given as, \(y \times I F=\int(Q \times I F) d x+C\) \(\therefore \quad y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C\) \(=\int(2 x \operatorname{sex}) d x+\int x^{2} \sec x \tan x d x+C\) \(\because \quad \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x\) Therefore, \(y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int_{0} x \sec x d x+C\) \(y \sec x=x^{2} \sec x+C \tag{i}\) \(\because \quad \mathrm{y}(0)=1 \Rightarrow 1(1)=0(1)+C\) \(\therefore \mathrm{C}=1\) \(\ddot{\text { Now, }} \mathrm{y}=\mathrm{x}^{2}+\cos \mathrm{x}\) from equation (i) and \(y^{\prime}=2 x-\sin x\) Now, \(\mathrm{y}^{\prime}\left(\frac{\pi}{4}\right)-\mathrm{y}^{\prime}\left(-\frac{\pi}{4}\right)=\left[2\left(\frac{\pi}{4}\right)-\frac{1}{\sqrt{2}}\right]-\left[2\left(-\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\right]=\pi-\sqrt{2}\)
Differential Equation
87480
If \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{\pi^{2}}{2 \sqrt{3}}\)
2 \(-\frac{\pi^{2}}{2 \sqrt{3}}\)
3 \(-\frac{\pi^{2}}{4 \sqrt{3}}\)
4 \(-\frac{\pi^{2}}{2}\)
Explanation:
(B) : Given, \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\) \(\Rightarrow \quad \frac{d y}{d x}-y \frac{\sin x}{\cos x}=\frac{6 x}{\cos x}\) \(\Rightarrow \quad \frac{d y}{d x}-y \tan x=\frac{6 x}{\cos x}\) Which is the linear differential equation of the form \(\frac{d y}{d x}+P x=Q\) So, \(\mathrm{IF}=\mathrm{e}^{-\int \tan x \mathrm{dx}}=\mathrm{e}^{-\log (\sec x)}=\cos \mathrm{x}\) \(\therefore \quad\) Required solution of differential equation is \(y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+c\) \(\because\) Given that, \(y\left(\frac{\pi}{3}\right)=0\) So, \(\quad 0=3\left(\frac{\pi}{3}\right)^{2}+\mathrm{C}\) \(\mathrm{C}=-\frac{\pi^{2}}{3}\) \(\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}\) Now, at \(\mathrm{x}=\frac{\pi}{6}\) \(y\left(\frac{\sqrt{3}}{2}\right)=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4}\) Hence, \(y=-\frac{\pi^{2}}{2 \sqrt{3}}\)
87477
Let \(y=y(x)\) be the solution of the differential equation, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\). If \(y\left(\frac{\pi}{2}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{4}{9 \sqrt{3}} \pi^{2}\)
2 \(\pi^{2}\)
3 \(-\frac{8}{9} \pi^{2}\)
4 \(-\frac{4}{9} \pi^{2}\)
Explanation:
(C) : Given, \(\sin x \frac{d y}{d x}+y \cos x=4 x, x \in(0, \pi)\) \(\therefore \quad \frac{d y}{d x}+y \frac{\cos x}{\sin x}=\frac{4 x}{\sin x}\) \(\frac{d y}{d x}+\cot x \cdot y=\frac{4 x}{\sin x}\) Equation (i) comparing with \(\frac{d y}{d x}+P(x) \cdot y=Q(x)\) \(\therefore \quad \mathrm{P}(\mathrm{x})=\cot \mathrm{x}\) and \(\mathrm{Q}(\mathrm{x})=\frac{4 \mathrm{x}}{\sin \mathrm{x}}\) I.F \(\quad=e^{\int P(x) d x}\) \(=\mathrm{e}^{\int \cot x \mathrm{dx}}\) \(=\mathrm{e}^{\int \log (|\sin x|) d x=\sin x}\) \(\therefore\) Solution is given by \(y \cdot \sin x=\int \frac{4 x}{\sin x} \cdot \sin x d x+c\) \(=\int 4 \mathrm{x} d \mathrm{x}+\mathrm{c}\) \(\begin{aligned} =\frac{4 \mathrm{x}^{2}}{2}+c \\ =2 \mathrm{x}^{2}+\mathrm{c}\end{aligned}\) \(\therefore \quad \mathrm{y} \sin \mathrm{x}=2 \mathrm{x}^{2}+\mathrm{c}\) \(\because\) Given \(y\left(\frac{\pi}{2}\right)=0 \quad\), when \(x=\frac{\pi}{2}, y=0\) \(\therefore\) Equation \(0=2 x\left(\frac{\pi}{2}\right)^{2}+\mathrm{c}\) \(\therefore \quad \mathrm{c}=\frac{\pi^{2}}{2}\) Put the value \(\mathrm{c}=\frac{\pi^{2}}{2}\) in equation (ii) \(y \sin x=2 x^{2}-\frac{\pi^{2}}{2}\) Find, \(y\left(\frac{\pi}{6}\right)\) \(\therefore \quad y \sin \frac{\pi}{6}=2\left(\frac{\pi}{6}\right)^{2}-\frac{\pi^{2}}{2}\) \(\mathrm{y} \times \frac{1}{2}=2 \cdot \frac{\pi^{2}}{36}-\frac{\pi^{2}}{2}\) \(\frac{y}{2}=\frac{\pi^{2}}{18}-\frac{\pi^{2}}{2}\) \(y=\frac{-8 \pi^{2}}{9}\) So, \(\quad y\left(\frac{\pi}{6}\right)=-\frac{8}{9} \pi^{2}\)
JEE Main-2018
Differential Equation
87478
Let \(y=y(x)\) be the solution of the differential equation, \(\quad\left(x^{2}+1\right)^{2} \frac{d y}{d x}+2 x\left(x^{2}+1\right) y=1\) such that \(y(0)=0\). If \(\sqrt{\mathrm{a}} y(1)=\frac{\pi}{32}\), then the value of ' \(a\) ' is
(D) : Given, \(\left(1+x^{2}\right) d t=\left(\tan ^{-1} x-t\right) d x\) \(\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\tan ^{-1} \mathrm{x}-\mathrm{t}}{1+\mathrm{x}^{2}}\) \(\frac{\mathrm{dt}}{\mathrm{dx}}+\frac{\mathrm{t}}{1+\mathrm{x}^{2}}=\frac{\tan ^{-1} \mathrm{x}}{1+\mathrm{x}^{2}}\) Which is form of \(\frac{d x}{d y}+P x=Q\) I.F \(=\mathrm{e}^{\int \frac{1}{1+\mathrm{x}^{2}} \mathrm{dx}} \quad\left(\because \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}\right)\) I.F. \(=\mathrm{e}^{\tan ^{-1} \mathrm{x}}\)
MHT CET-2020
Differential Equation
87479
Let \(y=y(x)\) be the solution of the differential equation, \(\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \quad\) such that \(y(0)=1\). Then
(A) : Given, \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ytan} x=2 \mathrm{x}+\mathrm{x}^{2} \tan x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) Linear differential equation in the form of \(\frac{d y}{d x}+P y=Q\) \(\therefore I F=\mathrm{e}^{\int \tan x \mathrm{dx}}=\mathrm{e}^{\log _{\mathrm{c}}(\sec \mathrm{x})}=\sec \mathrm{x}\) Now, solution of linear differential equation is given as, \(y \times I F=\int(Q \times I F) d x+C\) \(\therefore \quad y(\sec x)=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C\) \(=\int(2 x \operatorname{sex}) d x+\int x^{2} \sec x \tan x d x+C\) \(\because \quad \int x^{2} \sec x \tan x d x=x^{2} \sec x-\int(2 x \sec x) d x\) Therefore, \(y \sec x=2 \int x \sec x d x+x^{2} \sec x-2 \int_{0} x \sec x d x+C\) \(y \sec x=x^{2} \sec x+C \tag{i}\) \(\because \quad \mathrm{y}(0)=1 \Rightarrow 1(1)=0(1)+C\) \(\therefore \mathrm{C}=1\) \(\ddot{\text { Now, }} \mathrm{y}=\mathrm{x}^{2}+\cos \mathrm{x}\) from equation (i) and \(y^{\prime}=2 x-\sin x\) Now, \(\mathrm{y}^{\prime}\left(\frac{\pi}{4}\right)-\mathrm{y}^{\prime}\left(-\frac{\pi}{4}\right)=\left[2\left(\frac{\pi}{4}\right)-\frac{1}{\sqrt{2}}\right]-\left[2\left(-\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\right]=\pi-\sqrt{2}\)
Differential Equation
87480
If \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\), then \(y\left(\frac{\pi}{6}\right)\) is equal to
1 \(\frac{\pi^{2}}{2 \sqrt{3}}\)
2 \(-\frac{\pi^{2}}{2 \sqrt{3}}\)
3 \(-\frac{\pi^{2}}{4 \sqrt{3}}\)
4 \(-\frac{\pi^{2}}{2}\)
Explanation:
(B) : Given, \(\cos x \frac{d y}{d x}-y \sin x=6 x,\left(0\lt x\lt \frac{x}{2}\right)\) and \(y\left(\frac{\pi}{3}\right)=0\) \(\Rightarrow \quad \frac{d y}{d x}-y \frac{\sin x}{\cos x}=\frac{6 x}{\cos x}\) \(\Rightarrow \quad \frac{d y}{d x}-y \tan x=\frac{6 x}{\cos x}\) Which is the linear differential equation of the form \(\frac{d y}{d x}+P x=Q\) So, \(\mathrm{IF}=\mathrm{e}^{-\int \tan x \mathrm{dx}}=\mathrm{e}^{-\log (\sec x)}=\cos \mathrm{x}\) \(\therefore \quad\) Required solution of differential equation is \(y(\cos x)=\int(6 x) \frac{\cos x}{\cos x} d x+c\) \(\because\) Given that, \(y\left(\frac{\pi}{3}\right)=0\) So, \(\quad 0=3\left(\frac{\pi}{3}\right)^{2}+\mathrm{C}\) \(\mathrm{C}=-\frac{\pi^{2}}{3}\) \(\therefore \quad y(\cos x)=3 x^{2}-\frac{\pi^{2}}{3}\) Now, at \(\mathrm{x}=\frac{\pi}{6}\) \(y\left(\frac{\sqrt{3}}{2}\right)=3 \frac{\pi^{2}}{36}-\frac{\pi^{2}}{3}=-\frac{\pi^{2}}{4}\) Hence, \(y=-\frac{\pi^{2}}{2 \sqrt{3}}\)
