Question

Find the solution of IVP for the differential equation (x+2)^2e^y when y(1) =0 y'= dy/dx=(x+2)^2.e^y; y(1)=0

Answer 1

The solution is
`y\:=-\ln(-(x^(3))/(3)-4x-2x^(2)+(22)/(3))`

Explanation:

We need to find the solution of IVP for differential equation
`(dy)/(dx)=(x+2)^(2)e^(y)` when
`y(1)=0`

`\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}`

`\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)`

`\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}`

`(1)/(e^y)y'\:=\left(x+2\right)^2`

`N\left(y\right)\cdot y'\:=M\left(x\right)`

`N\left(y\right)=(1)/(e^y),\:\quad M\left(x\right)=\left(x+2\right)^2`

`\mathrm{Solve\:}\:(1)/(e^y)y'\:=\left(x+2\right)^2`

`(1)/(e^y)y'\:=x^(2)+4+4x`

Integrate both the sides with respect to dx

`\int(1)/(e^y)y'dx\:=\intx^(2)dx+4\int dx+4\int x dx`

`\int e^(-y)dy\:=\intx^(2)dx+4\int dx+4\int x dx`

`-(1)/(e^(y))\:=(x^(3))/(3)+4x+4(x^(2))/(2)+c_1`

`-(1)/(e^(y))\:=(x^(3))/(3)+4x+2x^(2)+c_1`

Since, IVP is y(1)=0

put x=1 and y=0 in above equation

`-(1)/(e^(0))\:=(1^(3))/(3)+4(1)+2(1)^(2)+c_1`

`-1\:=(1)/(3)+4+2+c_1`

`-1\:=(19)/(3)+c_1`

add both the sides by
`-(19)/(3)`

`-1-(19)/(3)\:=(19)/(3)-(19)/(3)+c_1`

`-1-(19)/(3)\:=c_1`

`-(22)/(3)\:=c_1`

so,

`-(1)/(e^(y))\:=(x^(3))/(3)+4x+2x^(2)-(22)/(3)`

Multiply both the sides by '-1'

`(1)/(e^(y))\:=-(x^(3))/(3)-4x-2x^(2)+(22)/(3)`

`e^(-y)\:=-(x^(3))/(3)-4x-2x^(2)+(22)/(3)`

Take natural logarithm both the sides,

`-y\:=\ln(-(x^(3))/(3)-4x-2x^(2)+(22)/(3))`

Multiply both the sides by '-1'

`y\:=-\ln(-(x^(3))/(3)-4x-2x^(2)+(22)/(3))`

Therefore, the solution is
`y\:=-\ln(-(x^(3))/(3)-4x-2x^(2)+(22)/(3))`