Solving an ordinary differential equation:Prove if y′(t)+3y(t)=6t+5,y(0)=3, then y(t)=2e−3t+2t+1.

Name: Solving an ordinary differential equation:Prove if y′(t)+3y(t)=6t+5,y(0)=3, then y(t)=2e−3t+2t+1.
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solving an ordinary differential equation:Prove if y′(t)+3y(t)=6t+5,y(0)=3, then y(t)=2e−3t+2t+1.

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stomachdm · Accepted Answer

The given equation is a first order differential equationy′(t)+3y(t)=6t+5can be solved by finding something called an integrating factor.For a first order differential equation of the formdydt+P(t)y=Q(t)The integrating factor is ef(t) wheref(t)=∫P(t)dtIn this case, the integrating factor is e3t because∫3dt=3tNow multiply the given equation throughout by e3t to gete3ty′(t)+3e3ty(t)=e3t(6t+5)The left hand side isddt(e3ty)Therefore ddt(e3ty)=e3t(6t+5)Integrate both sides now(e3ty)=∫6te3tdt+5∫e3tdte3ty=6te3t3−6∫e3t3dt+53e3t=6te3t3+e3t+constant=2te3t+e3t+constantBut y(0)=3 therefore the constant factor is 2e3ty(t)=2te3t+e3t+2⇒y(t)=2t+1+2e−3t

SlabydouluS62 · Accepted Answer

Set y(t)=2e−3t+2t+1+z(t) and substitute it to the equations:y′(t)+3y(t)=6t+5y(0)=3We get: −6e−3t+2+z′(t)+6e−3t+6t+3+3z(t)=6t+53+z(0)=3what simplifies to:z′(t)+3z(t)=0z(0)=0,but this simple ODE has only one solution, namely z(t)=0, and that completes the proof.

RizerMix · Accepted Answer

First solve the homogeneous equation y′+3y=0. This gives a homegeneous solution yh. It remains find to a particular solution yp. The solution to the ODE is then given by y=yh+yp. Since the non-homogeneous part of the equation is a polynomial of degree 1, try yp=t+c, where c is some constant. This constant is uniquely determined by the initial condition.

Solving an ordinary differential equation: Prove if y'(t)+3y(t)=6t+5, y(0)=3, then y(t)=2e^{-3t}+2t+1.

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