KCET · Maths · Differential Equations
If \(y=a \sin x+b \cos x\), then \(y^2+\left(\frac{d y}{d x}\right)^2\) is a
- A function of \(y\)
- B function of \(x\) and \(y\)
- C constant
- D function of \(x\)
Answer & Solution
Correct Answer
(C) constant
Step-by-step Solution
Detailed explanation
Given, \(y=a \sin x+b \cos x\)
\(\begin{aligned} & \frac{d y}{d y}=a \cos x-b \sin x \\ & \begin{aligned} y^2+\left(\frac{d y}{d x}\right)^2=(a \sin x+b \cos x)^2 \\ +(a \cos x-b \sin x)^2\end{aligned}\end{aligned}\)
\(\begin{aligned} & =a^2 \sin ^2 x+b^2 \cos ^2 x+2 a b \sin x \\ & \quad \cdot \cos x+a^2 \cos ^2 x+b^2 \sin ^2 x-2 a b \sin x \cdot \cos x \\ & =a^2\left(\sin ^2 x+\cos ^2 x\right)+b^2\left(\sin ^2 x+\cos ^2 x\right) \\ & =a^2+b^2 \quad\left[\because \sin ^2 x+\cos ^2 x=1\right] \\ & =\text { constant }\end{aligned}\)
\(\begin{aligned} & \frac{d y}{d y}=a \cos x-b \sin x \\ & \begin{aligned} y^2+\left(\frac{d y}{d x}\right)^2=(a \sin x+b \cos x)^2 \\ +(a \cos x-b \sin x)^2\end{aligned}\end{aligned}\)
\(\begin{aligned} & =a^2 \sin ^2 x+b^2 \cos ^2 x+2 a b \sin x \\ & \quad \cdot \cos x+a^2 \cos ^2 x+b^2 \sin ^2 x-2 a b \sin x \cdot \cos x \\ & =a^2\left(\sin ^2 x+\cos ^2 x\right)+b^2\left(\sin ^2 x+\cos ^2 x\right) \\ & =a^2+b^2 \quad\left[\because \sin ^2 x+\cos ^2 x=1\right] \\ & =\text { constant }\end{aligned}\)
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