MHT CET · Maths · Differential Equations
The differential equation whose solution is \(y=c_{1} \operatorname{cosax}+c_{2} \operatorname{sinax}\left(\right.\) Where \(c_{1}\)
and \(c_{2}\) are arbitrary constants) is
- A \(\frac{d^{2} y}{d x^{2}}-a^{2} y=0\)
- B \(\frac{d^{2} y}{d x^{2}}+a^{2} y=0\)
- C \(\frac{d^{2} y}{d x^{2}}+a y^{2}=0\)
- D \(\frac{d^{2} y}{d x^{2}}+y^{2}=0\)
Answer & Solution
Correct Answer
(B) \(\frac{d^{2} y}{d x^{2}}+a^{2} y=0\)
Step-by-step Solution
Detailed explanation
Given \(y=c_{1} \cos a x+c_{2} \sin a x\)
\(\therefore \frac{d y}{d x}=-a c_{1} \sin a x+a c_{2} \cos a x\)
\(\therefore y\left(\frac{1}{1-x^{2}}\right)=\int \frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \times \frac{1}{\left(1-x^{2}\right)} d x\)
\(=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)^{3 / 2}} d x=\left(-\frac{1}{2}\right) \frac{\left(1-x^{2}\right)^{\frac{1}{2}}}{\left(-\frac{1}{2}\right)}+c\)
\(y\left(\frac{1}{1-x^{2}}\right)=\frac{1}{\sqrt{1-x^{2}}}+c \Rightarrow y=\sqrt{1-x^{2}}~+\) \(c\left(1-x^{2}\right)\)
\(\therefore \frac{d y}{d x}=-a c_{1} \sin a x+a c_{2} \cos a x\)
\(\therefore y\left(\frac{1}{1-x^{2}}\right)=\int \frac{x}{\left(1-x^{2}\right)^{\frac{1}{2}}} \times \frac{1}{\left(1-x^{2}\right)} d x\)
\(=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)^{3 / 2}} d x=\left(-\frac{1}{2}\right) \frac{\left(1-x^{2}\right)^{\frac{1}{2}}}{\left(-\frac{1}{2}\right)}+c\)
\(y\left(\frac{1}{1-x^{2}}\right)=\frac{1}{\sqrt{1-x^{2}}}+c \Rightarrow y=\sqrt{1-x^{2}}~+\) \(c\left(1-x^{2}\right)\)
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