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Differential Equations question

2021 · 20 Jul · Shift 2 · Q33
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  5. /2021 · 20 Jul · Shift 2 · Q33

Differential Equations question

2021 · 20 Jul · Shift 2 · Q33

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) satisfies the equation dydx−∣A∣=0{{dy} \over {dx}} - |A| = 0dxdy​−∣A∣=0, for all x > 0, where A=[ysin⁡x10−11201x]A = \left[ {\begin{matrix} y & {\sin x} & 1 \\ 0 & { - 1} & 1 \\ 2 & 0 & {{1 \over x}} \\ \end{matrix} } \right]A=​y02​sinx−10​11x1​​​. If y(π)=π+2y(\pi ) = \pi + 2y(π)=π+2, then the value of y(π2)y\left( {{\pi \over 2}} \right)y(2π​) is :
  1. A
    π2+4π{\pi \over 2} + {4 \over \pi }2π​+π4​
  2. B
    π2−1π{\pi \over 2} - {1 \over \pi }2π​−π1​
  3. C
    3π2−1π{{3\pi } \over 2} - {1 \over \pi }23π​−π1​
  4. D
    π2−4π{\pi \over 2} - {4 \over \pi }2π​−π4​
View written solutionFree

Correct answer: A

  1. We are given the differential equation

dydx−∣A∣=0⇒dydx=∣A∣,\frac{dy}{dx}-|A|=0 \quad \Rightarrow \quad \frac{dy}{dx}=|A|,dxdy​−∣A∣=0⇒dxdy​=∣A∣,

where

y & \sin x & 1\\ 0 & -1 & 1\\ 2 & 0 & \frac1x \end{bmatrix}.$$ We need to compute $|A|$. --- 2. Find the determinant of $A$ by expanding along the first row: $$|A|=y\begin{vmatrix}-1 & 1\\ 0 & \frac1x\end{vmatrix}-\sin x\begin{vmatrix}0 & 1\\ 2 & \frac1x\end{vmatrix}+1\begin{vmatrix}0 & -1\\ 2 & 0\end{vmatrix}.$$ Now compute each minor: $$\begin{vmatrix}-1 & 1\\ 0 & \frac1x\end{vmatrix}=(-1)\cdot\frac1x-1\cdot 0=-\frac1x,$$ $$\begin{vmatrix}0 & 1\\ 2 & \frac1x\end{vmatrix}=0\cdot\frac1x-1\cdot 2=-2,$$ $$\begin{vmatrix}0 & -1\\ 2 & 0\end{vmatrix}=0\cdot 0-(-1)\cdot 2=2.$$ So, $$|A|=y\left(-\frac1x\right)-\sin x(-2)+1\cdot 2.$$ Hence, $$|A|=-\frac{y}{x}+2\sin x+2.$$ Therefore the differential equation becomes $$\frac{dy}{dx}=-\frac{y}{x}+2\sin x+2.$$ Rearrange: $$\frac{dy}{dx}+\frac{1}{x}y=2\sin x+2.$$ --- 3. This is a linear differential equation: $$\frac{dy}{dx}+P(x)y=Q(x), \quad P(x)=\frac1x.$$ Its integrating factor is $$\text{I.F.}=e^{\int \frac1x\,dx}=e^{\ln x}=x \qquad (x>0).$$ Multiply the equation by $x$: $$x\frac{dy}{dx}+y=2x\sin x+2x.$$ The left side is: $$\frac{d}{dx}(xy)=2x\sin x+2x.$$ Integrate both sides: $$xy=\int 2x\sin x\,dx+\int 2x\,dx+C.$$ --- 4. Compute the integrals. First, $$\int 2x\sin x\,dx=2\int x\sin x\,dx.$$ Using integration by parts: $$\int x\sin x\,dx=-x\cos x+\sin x.$$ Thus, $$\int 2x\sin x\,dx=2(-x\cos x+\sin x)=-2x\cos x+2\sin x.$$ Also, $$\int 2x\,dx=x^2.$$ So, $$xy=-2x\cos x+2\sin x+x^2+C.$$ Hence, $$y=-2\cos x+\frac{2\sin x}{x}+x+\frac{C}{x}.$$ --- 5. Use the condition $y(\pi)=\pi+2$. Substitute $x=\pi$: $$y(\pi)=-2\cos\pi+\frac{2\sin\pi}{\pi}+\pi+\frac{C}{\pi}.$$ Now, $$\cos\pi=-1, \qquad \sin\pi=0.$$ Therefore, $$y(\pi)=-2(-1)+0+\pi+\frac{C}{\pi}=2+\pi+\frac{C}{\pi}.$$ Given $y(\pi)=\pi+2$, we get $$\pi+2=\pi+2+\frac{C}{\pi} \Rightarrow \frac{C}{\pi}=0 \Rightarrow C=0.$$ So the solution is $$y=x-2\cos x+\frac{2\sin x}{x}.$$ --- 6. Now find $y\left(\frac\pi2\right)$: $$y\left(\frac\pi2\right)=\frac\pi2-2\cos\frac\pi2+\frac{2\sin(\pi/2)}{\pi/2}.$$ Using $$\cos\frac\pi2=0, \qquad \sin\frac\pi2=1,$$ we get $$y\left(\frac\pi2\right)=\frac\pi2+\frac{2}{\pi/2}=\frac\pi2+\frac{4}{\pi}.$$ --- 7. Compare with the options: $$\boxed{y\left(\frac\pi2\right)=\frac\pi2+\frac{4}{\pi}}.$$ So the correct option is **A**.
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