A) 5
B) 4
C) 3
D) 2
Correct Answer: C
Solution :
We have, \[y=({{C}_{1}}+{{C}_{2}})\cos (x+{{C}_{3}})-{{C}_{4}}{{e}^{x+{{C}_{5}}}}\] ? (i) \[\Rightarrow \]\[y=({{C}_{1}}+{{C}_{2}})\cos (x+{{C}_{3}})-{{C}_{4}}{{e}^{x}}\cdot {{e}^{{{C}_{5}}}}\] Now, let\[{{C}_{1}}+{{C}_{2}}=A,\,\,{{C}_{3}}=B,\,\,{{C}_{4}}{{e}^{{{C}_{5}}}}=C\] \[\Rightarrow \] \[y=A\cos (x+B)-C{{e}^{x}}\] ... (ii) On differentiating w.r.t. x, we get \[\frac{dy}{dx}=-A\sin (x+B)-C{{e}^{x}}\] ... (iii) Again, differentiating w.r.t.\[x\], we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-A\cos (x+B)-C{{e}^{x}}\] ... (iv) \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-y-2C{{e}^{x}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+y=-2C{{e}^{x}}\] ... (v) Again, differentiating w.r.t. x, we get \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+\frac{dy}{dx}=-2C{{e}^{x}}\] ... (vi) \[\Rightarrow \] \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+\frac{dy}{dx}=\frac{{{d}^{2}}y}{d{{x}^{2}}}+y\] [from Eq.(v)] which is a differential equation of order 3.
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