For what values of r does the function y=erx satisfy the differential equation y″−4y′+y=0?

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Vivian Soares · Accepted Answer

Step 11) Find y′ and y″2) Substitute the expressions for y, y′, y″ in the given differential equation3) Solve for rStep 2It is given thaty=erxDifferentiate with respect to xy′=d(erx)dxThe Chain Rule for differentiald[f[g(x)]]dx=d[f[g(x)]]d[g(x)]×d[g(x)]dxdydx=d(erx)d(rx)×d(rx)dxy′=erx×(r×1×x1−1)Remember that:1) the derivative of ex is ex itself2) the derivative of xn is nxn−1, this is known as power ruley′=rerxStep 3Note that:y′=ryDeffirntiate again, to gety″=ry′=r×rerx=r2erxSubstitute the expressions for y, y′, y″ in the following differential equationy″−4y′+y=0r2erx−4rerx+erx=0Since erx is never 0, we can divide both sides by itr2−4r+1=0add 3 to both sidesr2−4r+4=3Notice that the LHS is a perfect square(r−2)2=3Take the square root of both sidesr−2=±3Add 2 to both sidesr=2±3

Thomas White · Accepted Answer

Step 1Derivatives.y′=rerxy″=r2erxStep 2Substitute into differential equationr2erx−4rerx+erx=0Step 3Factorerx(r2−4r+1)=0Step 4Solve:erx=0 or r2−4r+1=0erx=0 does not existr2−4r+1=0 must be solved with quadratic formula.Step 5Quadratic formula:r=−−4±(−4)2−4×1×12×1r=4±16−42r=2±122r=2±3

nick1337 · Accepted Answer

Step 1y=1erxy′=rerxy″=r2erxr2−4r+1=0r=[4±16−4]2r=[4±23]2r=2±3

For what values of r does the function y =

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