Respuesta :
Answer:
The solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Step-by-step explanation:
We need to find the solution of IVP for differential equation [tex]\frac{dy}{dx}=(x+2)^{2}e^{y}[/tex] when [tex]y(1)=0[/tex]
[tex]\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}[/tex]
[tex]\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)[/tex]
[tex]\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}[/tex]
[tex]\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]
[tex]N\left(y\right)\cdot y'\:=M\left(x\right)[/tex]
[tex]N\left(y\right)=\frac{1}{e^y},\:\quad M\left(x\right)=\left(x+2\right)^2[/tex]
[tex]\mathrm{Solve\:}\:\frac{1}{e^y}y'\:=\left(x+2\right)^2[/tex]
[tex]\frac{1}{e^y}y'\:=x^{2}+4+4x[/tex]
Integrate both the sides with respect to dx
[tex]\int\frac{1}{e^y}y'dx\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]
[tex]\int e^{-y}dy\:=\intx^{2}dx+4\int dx+4\int x dx[/tex]
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+4\frac{x^{2}}{2}+c_1[/tex]
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}+c_1[/tex]
Since, IVP is y(1)=0
put x=1 and y=0 in above equation
[tex]-\frac{1}{e^{0}}\:=\frac{1^{3}}{3}+4(1)+2(1)^{2}+c_1[/tex]
[tex]-1\:=\frac{1}{3}+4+2+c_1[/tex]
[tex]-1\:=\frac{19}{3}+c_1[/tex]
add both the sides by [tex]-\frac{19}{3}[/tex]
[tex]-1-\frac{19}{3}\:=\frac{19}{3}-\frac{19}{3}+c_1[/tex]
[tex]-1-\frac{19}{3}\:=c_1[/tex]
[tex]-\frac{22}{3}\:=c_1[/tex]
so,
[tex]-\frac{1}{e^{y}}\:=\frac{x^{3}}{3}+4x+2x^{2}-\frac{22}{3}[/tex]
Multiply both the sides by '-1'
[tex]\frac{1}{e^{y}}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]
[tex]e^{-y}\:=-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3}[/tex]
Take natural logarithm both the sides,
[tex]-y\:=\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Multiply both the sides by '-1'
[tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
Therefore, the solution is [tex]y\:=-\ln(-\frac{x^{3}}{3}-4x-2x^{2}+\frac{22}{3})[/tex]
