For a=b=c=1 follows y'+y=e^x

$${y}^{\prime}\left(x\right)+ay\left(x\right)=c{e}^{bx}$$

The general solution of the first order differential equation with constant coefficients is:

$y\left(x\right)=\frac{c{e}^{bx}}{a+b}+k{e}^{-ax}$

In the first step the homogeneous equation has to be solved.

$$y\prime +ay=0$$

Solution of the homogeneous linear differential equation of first order with constant coefficients:

$y\prime =-ay$

Transformation of equation

$\frac{y\prime}{y}=-a$

Division by y

$\left(\mathrm{ln}y\right)\prime =-a$

Applying the chain rule

$\mathrm{ln}y=-a\int \mathrm{dx}=-ax+\stackrel{~}{k}$

Integration

${y}_{h}=k{e}^{-ax}$

General solution of the homogeneous equation with undetermined constants k

Variation of the constants:

In a second step the inhomogeneous differential equations can be obtained from the homogeneous one. Generally the solution of the inhomogeneous equation is given by the solution of the homogeneous equation plus a special solution of the inhomogeneous equation. The special solution can be obtained by the method of variation of constants. Here the constant k of the homogeneous solution is assumed as a function of x and the homogeneous solution is inserted into the inhomogeneous equation. k(x) is then determined so that the equation is fulfilled.

${y}_{h}^{\prime}=k\prime {e}^{-ax}-ak{e}^{-ax}$

Derivation of the homogeneous solution with k as a function of x

$k\prime {e}^{-ax}-ak{e}^{-ax}$$+ak{e}^{-ax}$$=c{e}^{bx}$

Insertion into the inhomogeneous equation

$k\prime =c{e}^{\left(a+b\right)x}$

By rearranging we obtain an equation for the determination of k

$k=\frac{c}{a+b}{e}^{\left(a+b\right)x}$

Integration gives k(x)

${y}_{s}=\frac{c}{a+b}{e}^{bx}$

Insertion of k(x) in y_{h} provides a special solution y_{s}

$y={y}_{s}+{y}_{h}=\frac{c}{a+b}{e}^{bx}+k{e}^{-ax}$

This is the general solution of the inhomogeneous differential equation

The calculator solves the initial value problem of y'+ay=ce^bx with the initial values x_{0}, y_{0}

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y'+ay=b y'+ay=ce^bx y'+2xy=xe^(-x^2) y'+xy=x y'+y=x y'=y^2 y'+y^2=1 y'=(Ay-a)(By-b)