-3-5-6--4 -3--1
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ANSWER
The particular solution is:
EXPLANATION
The given Ordinary Differential Equation is
The general solution to this Differential equation is:
To find the particular solution, we need to apply the initial conditions (ICs)
This implies that;
Hence the particular solution is
<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2><h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2><h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2><h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
