JEE MAIN - Mathematics (2024 - 8th April Morning Shift - No. 6)
Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
$$\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$$
$$\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0$$
Explanation
$$\int_\limits0^a f(x) d x=e^{-a}+4 a^2+a-1$$
Differentiate equation w.r.t. 'a'
$$\begin{aligned} & f(a)=-e^{-a}+8 a+1 \\ & \Rightarrow f(x)=-e^{-x}+8 x+1 \end{aligned}$$
And $$y=c_1 f(x)+c_2$$
$$\begin{aligned} & y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \\ & y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8} \end{aligned}$$
$$y^{\prime \prime}=-c_1 e^{-x}$$
put value of $$c_1$$
$$\begin{aligned} & \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \\ & \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1 \end{aligned}$$
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