Math, asked by aanchalyadav656, 6 months ago

find differential equations for y=Acosx-Bsinx

Answers

Answered by mxtridev

0

Answer:

d²y/dx² + y = 0

Step-by-step explanation:

y=Acosx-Bsinx

dy/dx= -Asinx-Bcosx

d²y/dx²= -Acosx+Bsinx= -[Acosx-Bsinx]= -y

then d²y/dx² + y = 0

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