87558
Let \(y=y(x)\) be the solution of the differential equation \(x d y=\left(y+x^{3} \cos x\right) d x\) with \(y(\pi)=0\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
2 \(\frac{\pi^{2}}{2}+\frac{\pi}{4}\)
3 \(\frac{\pi^{2}}{2}-\frac{\pi}{4}\)
4 \(\frac{\pi^{2}}{4}-\frac{\pi}{2}\)
Explanation:
(A) : Given, differential equation\(x d y=\left(y+x^{3} \cos x\right) d x\) \(x d y=y d y+x^{3} \cos x d x\) \(x d y-y d y=x^{3} \cos x d x\) \(\frac{x d y-y d y}{x^{2}}=x \cos x d x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)=\int \mathrm{x} \cos \mathrm{x} \mathrm{dx}\) \(\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x} \sin \mathrm{x}-\int 1 \cdot \sin \mathrm{xd} \mathrm{x}\) \(\underline{\mathrm{y}}=\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}+\mathrm{c}\) At, \(\quad \mathrm{x}=\pi\), and \(\mathrm{y}=0\) \(0=-1+\mathrm{c}\) On substituting the value of \(c\) is equation (i) we get \(\frac{y}{x}=x \sin x+\cos x+1\) \(y=x^{2} \sin x+x \cos x+x\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) So, \(\quad y\left(\frac{\pi}{2}\right)=\frac{\pi^{2}}{4}(1)+0+\frac{\pi}{2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
JEE Main 25.07.2021
Differential Equation
87559
Let \(y=y(x)\) be the solution of the differential equation \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1\) , with \(y(2)=-2\). Then \(y(3)\) is equal to
(B) : Given, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) On integrating both sides \(\Rightarrow \quad \frac{1}{2} \log |2 \mathrm{x}+3|=\frac{1}{2} \log |2 \mathrm{y}-1|+\log \mathrm{c}\) \(\Rightarrow \quad \log \left|\frac{2 x+3}{2 y-1}\right|=\log \mathrm{c}^{2}\) \(\Rightarrow \quad \frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{K}\), Where \(\mathrm{K}=\mathrm{c}^{2}\) So, option (b) is correct.
Manipal UGET-2017]**#
Differential Equation
87561
If the solution of the differential equation \(\frac{d y}{d x}=\frac{c y+3}{2 x+f}\), represents a circle, then the value of \(a\) is
1 2
2 -2
3 3
4 -4
Explanation:
(B) : \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{cy}+3}{2 \mathrm{x}+\mathrm{f}}\) \((a x+3) d x=(2 y+f) d y\) On integrating both side, \(a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) \(-\frac{a}{2} x^{2}+y^{2}-3 x+f y+c=0\) This will represent a circle if \(-\frac{a}{2}=1\) \(\left(\because\right.\) Coefficient of \(x^{2}=\) coefficient of \(\left.y^{2}\right)\) \(a=-2\)
87558
Let \(y=y(x)\) be the solution of the differential equation \(x d y=\left(y+x^{3} \cos x\right) d x\) with \(y(\pi)=0\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
2 \(\frac{\pi^{2}}{2}+\frac{\pi}{4}\)
3 \(\frac{\pi^{2}}{2}-\frac{\pi}{4}\)
4 \(\frac{\pi^{2}}{4}-\frac{\pi}{2}\)
Explanation:
(A) : Given, differential equation\(x d y=\left(y+x^{3} \cos x\right) d x\) \(x d y=y d y+x^{3} \cos x d x\) \(x d y-y d y=x^{3} \cos x d x\) \(\frac{x d y-y d y}{x^{2}}=x \cos x d x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)=\int \mathrm{x} \cos \mathrm{x} \mathrm{dx}\) \(\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x} \sin \mathrm{x}-\int 1 \cdot \sin \mathrm{xd} \mathrm{x}\) \(\underline{\mathrm{y}}=\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}+\mathrm{c}\) At, \(\quad \mathrm{x}=\pi\), and \(\mathrm{y}=0\) \(0=-1+\mathrm{c}\) On substituting the value of \(c\) is equation (i) we get \(\frac{y}{x}=x \sin x+\cos x+1\) \(y=x^{2} \sin x+x \cos x+x\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) So, \(\quad y\left(\frac{\pi}{2}\right)=\frac{\pi^{2}}{4}(1)+0+\frac{\pi}{2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
JEE Main 25.07.2021
Differential Equation
87559
Let \(y=y(x)\) be the solution of the differential equation \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1\) , with \(y(2)=-2\). Then \(y(3)\) is equal to
(B) : Given, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) On integrating both sides \(\Rightarrow \quad \frac{1}{2} \log |2 \mathrm{x}+3|=\frac{1}{2} \log |2 \mathrm{y}-1|+\log \mathrm{c}\) \(\Rightarrow \quad \log \left|\frac{2 x+3}{2 y-1}\right|=\log \mathrm{c}^{2}\) \(\Rightarrow \quad \frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{K}\), Where \(\mathrm{K}=\mathrm{c}^{2}\) So, option (b) is correct.
Manipal UGET-2017]**#
Differential Equation
87561
If the solution of the differential equation \(\frac{d y}{d x}=\frac{c y+3}{2 x+f}\), represents a circle, then the value of \(a\) is
1 2
2 -2
3 3
4 -4
Explanation:
(B) : \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{cy}+3}{2 \mathrm{x}+\mathrm{f}}\) \((a x+3) d x=(2 y+f) d y\) On integrating both side, \(a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) \(-\frac{a}{2} x^{2}+y^{2}-3 x+f y+c=0\) This will represent a circle if \(-\frac{a}{2}=1\) \(\left(\because\right.\) Coefficient of \(x^{2}=\) coefficient of \(\left.y^{2}\right)\) \(a=-2\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Differential Equation
87558
Let \(y=y(x)\) be the solution of the differential equation \(x d y=\left(y+x^{3} \cos x\right) d x\) with \(y(\pi)=0\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
2 \(\frac{\pi^{2}}{2}+\frac{\pi}{4}\)
3 \(\frac{\pi^{2}}{2}-\frac{\pi}{4}\)
4 \(\frac{\pi^{2}}{4}-\frac{\pi}{2}\)
Explanation:
(A) : Given, differential equation\(x d y=\left(y+x^{3} \cos x\right) d x\) \(x d y=y d y+x^{3} \cos x d x\) \(x d y-y d y=x^{3} \cos x d x\) \(\frac{x d y-y d y}{x^{2}}=x \cos x d x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)=\int \mathrm{x} \cos \mathrm{x} \mathrm{dx}\) \(\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x} \sin \mathrm{x}-\int 1 \cdot \sin \mathrm{xd} \mathrm{x}\) \(\underline{\mathrm{y}}=\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}+\mathrm{c}\) At, \(\quad \mathrm{x}=\pi\), and \(\mathrm{y}=0\) \(0=-1+\mathrm{c}\) On substituting the value of \(c\) is equation (i) we get \(\frac{y}{x}=x \sin x+\cos x+1\) \(y=x^{2} \sin x+x \cos x+x\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) So, \(\quad y\left(\frac{\pi}{2}\right)=\frac{\pi^{2}}{4}(1)+0+\frac{\pi}{2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
JEE Main 25.07.2021
Differential Equation
87559
Let \(y=y(x)\) be the solution of the differential equation \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1\) , with \(y(2)=-2\). Then \(y(3)\) is equal to
(B) : Given, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) On integrating both sides \(\Rightarrow \quad \frac{1}{2} \log |2 \mathrm{x}+3|=\frac{1}{2} \log |2 \mathrm{y}-1|+\log \mathrm{c}\) \(\Rightarrow \quad \log \left|\frac{2 x+3}{2 y-1}\right|=\log \mathrm{c}^{2}\) \(\Rightarrow \quad \frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{K}\), Where \(\mathrm{K}=\mathrm{c}^{2}\) So, option (b) is correct.
Manipal UGET-2017]**#
Differential Equation
87561
If the solution of the differential equation \(\frac{d y}{d x}=\frac{c y+3}{2 x+f}\), represents a circle, then the value of \(a\) is
1 2
2 -2
3 3
4 -4
Explanation:
(B) : \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{cy}+3}{2 \mathrm{x}+\mathrm{f}}\) \((a x+3) d x=(2 y+f) d y\) On integrating both side, \(a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) \(-\frac{a}{2} x^{2}+y^{2}-3 x+f y+c=0\) This will represent a circle if \(-\frac{a}{2}=1\) \(\left(\because\right.\) Coefficient of \(x^{2}=\) coefficient of \(\left.y^{2}\right)\) \(a=-2\)
87558
Let \(y=y(x)\) be the solution of the differential equation \(x d y=\left(y+x^{3} \cos x\right) d x\) with \(y(\pi)=0\), then \(y\left(\frac{\pi}{2}\right)\) is equal to
1 \(\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
2 \(\frac{\pi^{2}}{2}+\frac{\pi}{4}\)
3 \(\frac{\pi^{2}}{2}-\frac{\pi}{4}\)
4 \(\frac{\pi^{2}}{4}-\frac{\pi}{2}\)
Explanation:
(A) : Given, differential equation\(x d y=\left(y+x^{3} \cos x\right) d x\) \(x d y=y d y+x^{3} \cos x d x\) \(x d y-y d y=x^{3} \cos x d x\) \(\frac{x d y-y d y}{x^{2}}=x \cos x d x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)=\int \mathrm{x} \cos \mathrm{x} \mathrm{dx}\) \(\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x} \sin \mathrm{x}-\int 1 \cdot \sin \mathrm{xd} \mathrm{x}\) \(\underline{\mathrm{y}}=\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}+\mathrm{c}\) At, \(\quad \mathrm{x}=\pi\), and \(\mathrm{y}=0\) \(0=-1+\mathrm{c}\) On substituting the value of \(c\) is equation (i) we get \(\frac{y}{x}=x \sin x+\cos x+1\) \(y=x^{2} \sin x+x \cos x+x\) At, \(\quad \mathrm{x}=\frac{\pi}{2}\) So, \(\quad y\left(\frac{\pi}{2}\right)=\frac{\pi^{2}}{4}(1)+0+\frac{\pi}{2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}\)
JEE Main 25.07.2021
Differential Equation
87559
Let \(y=y(x)\) be the solution of the differential equation \(x\left(1-x^{2}\right) \frac{d y}{d x}+\left(3 x^{2} y-y-4 x^{3}\right)=0, x>1\) , with \(y(2)=-2\). Then \(y(3)\) is equal to
(B) : Given, \((2 y-1) d x-(2 x+3) d y=0\) \((2 y-1) d x=(2 x+3) d y\) \(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) On integrating both sides \(\Rightarrow \quad \frac{1}{2} \log |2 \mathrm{x}+3|=\frac{1}{2} \log |2 \mathrm{y}-1|+\log \mathrm{c}\) \(\Rightarrow \quad \log \left|\frac{2 x+3}{2 y-1}\right|=\log \mathrm{c}^{2}\) \(\Rightarrow \quad \frac{2 \mathrm{x}+3}{2 \mathrm{y}-1}=\mathrm{K}\), Where \(\mathrm{K}=\mathrm{c}^{2}\) So, option (b) is correct.
Manipal UGET-2017]**#
Differential Equation
87561
If the solution of the differential equation \(\frac{d y}{d x}=\frac{c y+3}{2 x+f}\), represents a circle, then the value of \(a\) is
1 2
2 -2
3 3
4 -4
Explanation:
(B) : \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{cy}+3}{2 \mathrm{x}+\mathrm{f}}\) \((a x+3) d x=(2 y+f) d y\) On integrating both side, \(a \frac{x^{2}}{2}+3 x=y^{2}+f y+c\) \(-\frac{a}{2} x^{2}+y^{2}-3 x+f y+c=0\) This will represent a circle if \(-\frac{a}{2}=1\) \(\left(\because\right.\) Coefficient of \(x^{2}=\) coefficient of \(\left.y^{2}\right)\) \(a=-2\)