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If \(f(x), f^{\prime}(x) f^{\prime \prime}(x)\), are positive functions and \(f(0)=1, f^{\prime}(0)=2\), then the solution of the differential equation \(\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0\) is
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Verified Answer
The correct answer is:
\(e^{2 x}\)
\(\begin{aligned}
& \text {Since, }\left|\begin{array}{ll}
f(x) & f^{\prime}(x) \\
f^{\prime}(x) & f^{\prime \prime}(x)
\end{array}\right|=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^2 \\
& \Rightarrow \quad \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\
\end{aligned}\)
On, integrating both sides, we get
\(\begin{aligned}
& \int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} d x=\int \frac{f^{\prime}(x)}{f(x)} d x \\
& \Rightarrow \quad \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+c \\
& \because \quad f(0)=1 \text { and } f^{\prime}(0)=2 \\
& \therefore \quad c=\log _e 2 \\
& \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+\log _e 2 \\
& \Rightarrow \quad f^{\prime}(x)=2 f(x) \\
& \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2 \\
& \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\int 2 d x \Rightarrow \log _e f(x)=2 x+c \\
& \because \quad f(0)=1 \text {, so } c=0 \\
& \because \quad \log _e f(x)=2 x \Rightarrow f(x)=e^{2 x} \\
\end{aligned}\)
Hence, option (1) is correct.
& \text {Since, }\left|\begin{array}{ll}
f(x) & f^{\prime}(x) \\
f^{\prime}(x) & f^{\prime \prime}(x)
\end{array}\right|=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)-\left(f^{\prime}(x)\right)^2=0 \\
& \Rightarrow \quad f(x) f^{\prime \prime}(x)=\left(f^{\prime}(x)\right)^2 \\
& \Rightarrow \quad \frac{f^{\prime \prime}(x)}{f^{\prime}(x)}=\frac{f^{\prime}(x)}{f(x)} \\
\end{aligned}\)
On, integrating both sides, we get
\(\begin{aligned}
& \int \frac{f^{\prime \prime}(x)}{f^{\prime}(x)} d x=\int \frac{f^{\prime}(x)}{f(x)} d x \\
& \Rightarrow \quad \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+c \\
& \because \quad f(0)=1 \text { and } f^{\prime}(0)=2 \\
& \therefore \quad c=\log _e 2 \\
& \log _e\left(f^{\prime}(x)\right)=\log _e(f(x))+\log _e 2 \\
& \Rightarrow \quad f^{\prime}(x)=2 f(x) \\
& \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=2 \\
& \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\int 2 d x \Rightarrow \log _e f(x)=2 x+c \\
& \because \quad f(0)=1 \text {, so } c=0 \\
& \because \quad \log _e f(x)=2 x \Rightarrow f(x)=e^{2 x} \\
\end{aligned}\)
Hence, option (1) is correct.
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