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find d²y/dx²,if y³+3ax²+x³=0 Related: Standard forms of Integration, ...
To find d²y/dx², we need to differentiate y twice with respect to x.
First, we use the product rule to differentiate y³ 3ax² x³=0:
y³ d/dx(3ax² x³) + 3ax² x³ d/dx(y³) = 0
Simplifying:
3y²(6ax^4) + 3ax²x³(3y² dy/dx) = 0
Now, we can solve for dy/dx:
dy/dx = -2ax / y
To find d²y/dx², we differentiate again using the quotient rule:
d/dx(dy/dx) = d/dx(-2ax/y)
= (-2a(dy/dx)(dy/dx)/y²) - (-2ax(d²y/dx²)/y)
Substituting in our expression for dy/dx:
d/dx(dy/dx) = (-2a(-2ax/y)(-2ax/y)/y²) - (-2ax(d²y/dx²)/y)
Simplifying:
d²y/dx² = (4a²x²/y²) - (2ax/y)(dy/dx)
Substituting in our expression for dy/dx:
d²y/dx² = (4a²x²/y²) + (4a²x²/y²)
= 8a²x²/y²
Therefore, d²y/dx² = 8a²x²/y².
First, we use the product rule to differentiate y³ 3ax² x³=0:
y³ d/dx(3ax² x³) + 3ax² x³ d/dx(y³) = 0
Simplifying:
3y²(6ax^4) + 3ax²x³(3y² dy/dx) = 0
Now, we can solve for dy/dx:
dy/dx = -2ax / y
To find d²y/dx², we differentiate again using the quotient rule:
d/dx(dy/dx) = d/dx(-2ax/y)
= (-2a(dy/dx)(dy/dx)/y²) - (-2ax(d²y/dx²)/y)
Substituting in our expression for dy/dx:
d/dx(dy/dx) = (-2a(-2ax/y)(-2ax/y)/y²) - (-2ax(d²y/dx²)/y)
Simplifying:
d²y/dx² = (4a²x²/y²) - (2ax/y)(dy/dx)
Substituting in our expression for dy/dx:
d²y/dx² = (4a²x²/y²) + (4a²x²/y²)
= 8a²x²/y²
Therefore, d²y/dx² = 8a²x²/y².
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