The given differential equation is $ (x^2+1)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0$. This is a second-order linear homogeneous differential equation. We are given that $y_1(x) = x$ is one solution. To find another linearly independent solution, we can use the method of reduction of order. Let the second solution be $y_2(x) = v(x) y_1(x) = vx$. We need to find the derivatives of $y_2$: $\frac{dy_2}{dx} = v + x\frac{dv}{dx}$ $\frac{d^2y_2}{dx^2} = \frac{dv}{dx} + \frac{dv} {dx} + x\frac{d^2v}{dx^2} = 2\frac{dv}{dx} + x\frac{d^2v}{dx^2}$Substitute these into the differential equation: $(x^2+1)\left(2\frac{dv}{dx} + x\frac{d^2v}{dx^2}\right) - 2x\left(v + x\frac{dv} {dx}\right) + 2(vx) = 0$Expand and group terms: $2(x^2+1)\frac{dv}{dx} + x(x^2+1)\frac{d^2v} {dx^2} - 2xv - 2x^2\frac{dv}{dx} + 2vx = 0$Simplify: $x(x^2+1)\frac{d^2v}{dx^2} + (2x^2+2 - 2x^2)\frac{dv}{dx} + (-2xv + 2vx) = 0$ $x(x^2+1)\frac{d^2v}{dx^2} + 2\frac{dv}{dx} = 0$Let $w = \frac{dv}{dx}$. Then $\frac{dw}{dx} = \frac{d^2v}{dx^2}$. The equation becomes: $x(x^2+1)w' + 2w = 0$ This is a first-order separable differential equation for $w$: $\frac{w'}{w} = -\frac{2}{x(x^2+1)}$ $\frac{1}{w} \frac{dw}{dx} = -\frac{2}{x(x^2+1)}$Integrate both sides with respect to $x$: $\int \frac{dw}{w} = -\int \frac{2} {x(x^2+1)} dx$For the integral on the right, use partial fraction decomposition: $\frac{2} {x(x^2+1)} = \frac{A}{x} + \frac{Bx+C}{x^2+1}$ $2 = A(x^2+1) + (Bx+C)x$ $2 = Ax^2 + A + Bx^2 + Cx$ $2 = (A+B)x^2 + Cx + A$Comparing coefficients: $x^2$: $A+B = 0 \implies B = -A$ $x$: $C = 0$ Constant: $A = 2$So, $A=2$, $B=-2$, $C=0$. $\frac{2}{x(x^2+1)} = \frac{2}{x} - \frac{2x} {x^2+1}$Now, integrate: $\int \frac{2}{x(x^2+1)} dx = \int \left(\frac{2}{x} - \frac{2x}{x^2+1}\right) dx$ $= 2 \ln|x| - \ln(x^2+1) + K_1$ $= \ln(x^2) - \ln(x^2+1) + K_1$ $= \ln\left(\frac{x^2} {x^2+1}\right) + K_1$So, $\ln|w| = - \ln\left(\frac{x^2}{x^2+1}\right) + K_2$ $\ln|w| = \ln\left(\frac{x^2+1}{x^2}\right) + K_2$ $w = e^{K_2} \frac{x^2+1}{x^2}$ Let $C_3 = e^{K_2}$. $w = C_3 \left(1 + \frac{1}{x^2}\right) $Now, $w = \frac{dv}{dx}$, so we integrate $w$ to find $v$: $v = \int C_3 \left(1 + \frac{1} {x^2}\right) dx$ $v = C_3 \left(x - \frac{1}{x} \right) + C_4$The general solution is $y = y_1 v = x \left(C_3 \left(x - \frac{1}{x}\right) + C_4\right)$ $y = C_3 x \left(x - \frac{1}{x}\right) + C_4 x$ $y = C_3 (x^2 - 1) + C_4 x$The two linearly independent solutions are $x^2-1$ and $x$. Since $y_1 = x$ is given, another linearly independent solution is $y_2 = x^2-1$. This corresponds to Option B.We can also check the options directly. For $y = \frac{1}{2}x$ (Option A), this is a multiple of $y_1=x$, so it's linearly dependent. For $y = \pi x$ (Option C), this is a multiple of $y_1=x$, so it's linearly dependent. For $y = 2x$ (Option D), this is a multiple of $y_1=x$, so it's linearly dependent.Therefore, $y=x^2-1$ is the only option that is not a trivial multiple of $y_1=x$. Let's verify $y=x^2-1$ in the differential equation. $y = x^2-1$ $\frac{dy}{dx} = 2x$ $\frac{d^2y}{dx^2} = 2$Substitute into $(x^2+1)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0$: $(x^2+1)(2) - 2x(2x) + 2(x^2-1) = 0$ $2x^2 + 2 - 4x^2 + 2x^2 - 2 = 0$ $(2x^2 - 4x^2 + 2x^2) + (2 - 2) = 0$ $0 + 0 = 0$ The equation holds true.
If y = x is one of the solution of differential equation $(x^2+1)\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = 0$, then another linearly independent solution is - - (1) $y=\frac{1}{2}x$ - (3) $y=\pi x$ - (5) Question not attempted
Asked in Asst. Statistical Officer (Agri. Deptt.) 2024 · Topic: Computer and Information Technology (Computational Physics: Numerical methods and algorithms for physical problems)
More Physics questions
- Which of the following statement is not correct? (1) The closed half spaces $H_1 = \{x | cx \ge z\}$ and $H_2 = \{x | cx \le z\}$ are convex sets. (2) Intersection of two convex sets is also a convex set. (3) Let $A = \{x_1 - x_2 | x_1 \in S, x_2 \in T\}$, where S and T are two non-empty disjoint convex sets, then A is a convex set. (4) If S and T are any two convex sets in $\mathbb{R} ^n$, then for all scalars $\alpha$ and $\beta$, the set $\alpha S + \beta T$ is not convex. (5) Question not attempted
- One independent solution of partial differential equation $yzp + zxq = xy$, (where $p = \frac{\partial z}{\partial x}$, $q = \frac{\partial z}{\partial y}$) is -
- If $u = \sin^{-1}\left(\frac{x^3+y^3}{\sqrt{x}+ \sqrt{y}}\right)$, then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is equal to -
- The hypergeometric function $_{2}F_{1} (\alpha, \beta; \beta; x)$ is equal to -
- Solution of the differential equation x $ \frac{d^2y}{dx^2} - (2x-1)\frac{dy}{dx} + (x-1)y = 0$ is - - (1) $y = \log_e x (C_1 e^x + C_2)$ - (2) $y = e^x (C_1 \log_e x + C_2)$ - (3) $y = \log_e x (C_1 e^{-x} + C_2)$ - (4) $y = e^{-x} (C_1 \log_e x + C_2)$ - (5) Question not attempted
- Lagrange's interpolation formula can be used - - (5) Question not attempted
- The value of f(1.22) from the following table using Stirling's Interpolation formula is - x : 1.0 1.1 1.2 1.3 1.4 - (5) Question not attempted
- If $J_n(x)$ denotes the Bessel's function of first kind of order n in argument x, then $J'_0(x)$ is equal to, where $J'_0(x)$ is the first derivative of $J_0(x)$ w.r.t. x - - (5) Question not attempted