If y=3e2x+2e3x, prove that dx2d2y−5dxdy+6y=0.
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Step-by-Step Solution
Step 1: Find the first derivative
To prove the given equation, we first need to find the first and second derivatives of y with respect to x. We start by differentiating the given function y=3e2x+2e3x once.
Step 2: Calculate the first derivative
Using the chain rule, the derivative of eax is aeax. Applying this, the derivative of 3e2x is 3⋅2e2x=6e2x, and the derivative of 2e3x is 2⋅3e3x=6e3x.
Step 3: Find the second derivative
Now we differentiate the first derivative, 6e2x+6e3x, to find the second derivative of y with respect to x.
Step 4: Calculate the second derivative
Again, applying the chain rule, the derivative of 6e2x is 6⋅2e2x=12e2x, and the derivative of 6e3x is 6⋅3e3x=18e3x.
Step 5: Substitute into the equation
Now we substitute the expressions for y, dxdy, and dx2d2y into the given equation dx2d2y−5dxdy+6y=0.
Step 6: Simplify the expression
Distribute the coefficients and expand the terms. We will then group like terms (e2x terms and e3x terms) to simplify the expression.
Step 7: Combine like terms
Combine the coefficients for e2x and e3x terms. We find that (12−30+18)=0 and (18−30+12)=0, which proves the equation is equal to zero.