Variation of parameters
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}={\left.{d}{x}\right.}\)
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
\(\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=\int{\left.{d}{x}\right.}\)
which is\(\displaystyle \ln{{\left|{{y}}\right|}}={x}+{c}\)
By taking exponents, we obtain
\(\displaystyle{\left|{{y}}\right|}={e}^{{{x}+{c}}}={e}^{x}\cdot{e}^{c}\)
Hence,we obtain
\(\displaystyle{y}={C}{e}^{x}\)
where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{x}\) is the compleentary solution
Next, we need to find the particular solution \(y_p\).
Therefore, we consider \(uy_c\), and try to find u, a function of x, that will make this work.
Let's assume that \(uy_c\). is a solution of the given equation. Hence, it satisfies the given equation. Substituting \(uy_c\), and its derivative in the equation gives
\(\displaystyle{\left({u}{y}_{{c}}\right)}'-{u}{y}_{{c}}={4}{e}^{x}\)
\(\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}-{u}{y}_{{c}}={4}{e}^{x}\)
\(\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}-{y}_{{c}}\right)}}}_{{\text{=0 }\ }}={4}{e}^{x}\)
since y_c is a solution
Therefore
\(\displaystyle{u}'{y}_{{c}}={4}{e}^{x}\Rightarrow{u}'=\frac{{{4}{e}^{x}}}{{{y}_{{c}}}}\)
which gives
\(\displaystyle{u}=\int\frac{{{e}^{x}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}\)
Now, we can find the function u :
\(\displaystyle{u}=\int\frac{{{4}{e}^{x}}}{{{e}^{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={4}\int{\left.{d}{x}\right.}\)
\(=4x+c\)
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
\(u=4x\)
Recall that \(y_p = uy_c\). Therefore,
\(\displaystyle{y}_{{p}}={4}{x}{e}^{x}\)
The general solution is
\(y=Cy_c+y_p\)
\(\displaystyle={e}^{x}{\left({C}+{4}{x}\right)}\)
Integrating Factor technique
This equation is linear with \(\displaystyle{P}{\left({x}\right)}=-{1}{\quad\text{and}\quad}{Q}{\left({x}\right)}={4}{e}^{x}\)
Hence,
\(\displaystyle{h}=\int{P}{\left.{d}{x}\right.}=-\int{\left.{d}{x}\right.}=-{x}\)
So, an integrating factor is
\(\displaystyle{e}^{h}={e}^{{-{x}}}\)
and the general solution is
\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+\int{4}{e}^{x}\cdot{e}^{{-{x}}}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+{4}\int{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+{4}{x}\right)}\)
Now, we can use the given initial condition. Substitute 0 for x and 4 for y to obtain the numeric value of c.
\(\displaystyle{4}={y}{\left({0}\right)}={e}^{0}{\left({c}+{4}\cdot{0}\right)}\Rightarrow{4}={c}\)
Hence, the solution is
\(\displaystyle{y}={e}^{x}{\left({c}+{4}{x}\right)}\)
First, solve the linear homogeneous equation by separating variables. Rearranging terms in the equation gives
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={y}\Leftrightarrow\frac{{{\left.{d}{y}\right.}}}{{y}}={\left.{d}{x}\right.}\)
Now, the variables are separated, x appears only on the right side, and y only on the left.
Integrate the left side in relation to y, and the right side in relation to x
\(\displaystyle\int\frac{{{\left.{d}{y}\right.}}}{{y}}=\int{\left.{d}{x}\right.}\)
which is\(\displaystyle \ln{{\left|{{y}}\right|}}={x}+{c}\)
By taking exponents, we obtain
\(\displaystyle{\left|{{y}}\right|}={e}^{{{x}+{c}}}={e}^{x}\cdot{e}^{c}\)
Hence,we obtain
\(\displaystyle{y}={C}{e}^{x}\)
where \(\displaystyle{C}\:=\pm{e}^{c}{\quad\text{and}\quad}{y}_{{c}}={e}^{x}\) is the compleentary solution
Next, we need to find the particular solution \(y_p\).
Therefore, we consider \(uy_c\), and try to find u, a function of x, that will make this work.
Let's assume that \(uy_c\). is a solution of the given equation. Hence, it satisfies the given equation. Substituting \(uy_c\), and its derivative in the equation gives
\(\displaystyle{\left({u}{y}_{{c}}\right)}'-{u}{y}_{{c}}={4}{e}^{x}\)
\(\displaystyle{u}'{y}_{{c}}+{u}{y}'_{{c}}-{u}{y}_{{c}}={4}{e}^{x}\)
\(\displaystyle{u}'{y}_{{c}}+\underbrace{{{u}{\left({y}'_{{c}}-{y}_{{c}}\right)}}}_{{\text{=0 }\ }}={4}{e}^{x}\)
since y_c is a solution
Therefore
\(\displaystyle{u}'{y}_{{c}}={4}{e}^{x}\Rightarrow{u}'=\frac{{{4}{e}^{x}}}{{{y}_{{c}}}}\)
which gives
\(\displaystyle{u}=\int\frac{{{e}^{x}}}{{{y}_{{c}}}}{\left.{d}{x}\right.}\)
Now, we can find the function u :
\(\displaystyle{u}=\int\frac{{{4}{e}^{x}}}{{{e}^{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={4}\int{\left.{d}{x}\right.}\)
\(=4x+c\)
Since we need to find only one function that will male this work, we don’t need to introduce the constant of integration c. Hence,
\(u=4x\)
Recall that \(y_p = uy_c\). Therefore,
\(\displaystyle{y}_{{p}}={4}{x}{e}^{x}\)
The general solution is
\(y=Cy_c+y_p\)
\(\displaystyle={e}^{x}{\left({C}+{4}{x}\right)}\)
Integrating Factor technique
This equation is linear with \(\displaystyle{P}{\left({x}\right)}=-{1}{\quad\text{and}\quad}{Q}{\left({x}\right)}={4}{e}^{x}\)
Hence,
\(\displaystyle{h}=\int{P}{\left.{d}{x}\right.}=-\int{\left.{d}{x}\right.}=-{x}\)
So, an integrating factor is
\(\displaystyle{e}^{h}={e}^{{-{x}}}\)
and the general solution is
\(\displaystyle{y}{\left({x}\right)}={e}^{{-{h}}}{\left({c}+\int{Q}{e}^{h}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+\int{4}{e}^{x}\cdot{e}^{{-{x}}}{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+{4}\int{\left.{d}{x}\right.}\right)}\)
\(\displaystyle={e}^{x}{\left({c}+{4}{x}\right)}\)
Now, we can use the given initial condition. Substitute 0 for x and 4 for y to obtain the numeric value of c.
\(\displaystyle{4}={y}{\left({0}\right)}={e}^{0}{\left({c}+{4}\cdot{0}\right)}\Rightarrow{4}={c}\)
Hence, the solution is
\(\displaystyle{y}={e}^{x}{\left({c}+{4}{x}\right)}\)
