JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Differential Equations

More JEE Main questions on Differential Equations

1. Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right)dy+\left(y-{\tan }^{-1}x\right)dx=0$, $y(0)=1$. Then the value of $y(1)$ is :
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2. Let $y=y(x)$ be the solution of the differential equation $\sec x\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=2+3\sin \mathrm{x},\mathrm{x}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, $y(0)=-\frac{7}{4}$. Then $y\left(\frac{\pi }{6}\right)$ is equal to:
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3. Let the solution curve of the differential equation $xdy-ydx=\sqrt{x^2+y^2}dx,x>0,y(1)=0$, be $y=y(x)$. Then $y(3)$ is equal to
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4. If $y=y(x)$ satisfies the differential equation
   $16(\sqrt{x+9\sqrt{x}})(4+\sqrt{9+\sqrt{x}})\cos ydy=(1+2\sin y)dx$, $x>0$ and $y(256)=\frac{\pi }{2}$, $y(49)=\alpha$, then $2\sin \alpha$ is equal to :
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5. Let $y=y(x)$ be the solution of the differential equation
   $x^{4}dy+\left(4x^{3}y+2\sin x\right)dx=0$, $x>0$, $y\left(\frac{\pi }{2}\right)=0$.
   
   Then ${\pi }^4\mathrm{y}\left(\frac{\pi }{3}\right)$ is equal to :
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6. If the solution curve $y=f(x)$ of the differential equation $\left(x^2-4\right)y^{\prime }-2xy+2x{\left(4-x^2\right)}^2=0,x>2$, passes through the point (3,15), then the local maximum value of f is
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