Solve that initial value problem of y’ + 2y = 3, y(0) = 1.

Question

Solve that initial value problem of y&#8217; + 2y = 3, y(0) = 1.

Suresh Chand · Accepted Answer

Definition of Initial Value Problem: A differential equation together with initial condition(s) is called the initial value problem. For example, (frac{d^2y}{d^2x} = 2frac{dy}{dx} + y = 0, y(0) = -1, y^{'}(0) = 1) Here (y(0) = -1, y^{'}(0) = 1) is an initial conditions   Problem Part: Solution: Given that y' + 2y = 1          ....... (i) y(0) = 1               .......(ii) Since (i) is linear differential equation of first order. Comparing (i) with y' + Py = Q then P = 2, Q = 3 Then (int P dx = 2 int dx = 2x) Here, the integrating factor (I.F) of (i) is (I.F. = e^{int Pdx} = e^{2x}) Now, Multiplying (i) by I.F. and then integrating we get, (y times I.F. = int(Q times I.F.)dx + C) (i.e enspace ye^{2x} = int (3) (e^{2x}) dx + C = 3 frac{e^{2x}}{2} + C)           .....(iii) By (ii), we have y(0) = 1, then (iii) becomes, (1 times e ^ {2 times 0} = frac{3}{2} e ^ {2 times 0} + C) (1 times 1 = frac{3}{2} times 1 + C) (C = -frac{1}{2}) Applying this value, We get (ye^{2x} = frac{e^{2x}}{2} - frac{1}{2}) (Rightarrow y = frac{1}{2} (1 + e^{-2x})) This is the solution of (i) that stratifies (ii).

Define initial value problem. Solve that initial value problem of y’ + 2y = 3, y(0) = 1.

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Define initial value problem. Solve that initial value problem of y’ + 2y = 3, y(0) = 1.

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