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Solve the given initial value problem.$6 x^{2} y^{\prime \prime}+7 x y^{\prime}-2 y=0, \quad y(1)=0, y^{\prime}(1)=1$

Calculus 3

Chapter 17

Second-Order Differential Equations

Section 4

Euler Equations

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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Solve the given initial va…

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Solve the given initial-va…

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Find the solution of the g…

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Find a solution to the ini…

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all right. Number 26 6 X squared Y double prime for a while if one equals zero. And why private one Because one So finding the our form we get six r squared plus ar minus two equals zero. Factoring we get to our minus 13 R plus two equals zero. So articles one half or negative two thirds. Therefore, a wise you go to see one route X plus C two over ext are two or three. Why primes See one over to root X minus two C 2/3 X to the five or three. Now, using the initial conditions, we find that C one plus C two equals zero and half see one minus two thirds C two equals one. Solving we got C one equals 6/7. C two equals *** 6/7. Therefore, our final equation is 6/7 squared X minus 6/7 x par three over to

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