Question

Find an explicit solution of the given initial-value problem

dy/dx = ye^(-x^2), y(5) = 1

Answer 1

$\frac{dy}{dx} = ye^{-x^2}$

$\frac{dy}{y} = e^{-x^2}dx$

$\int\frac{du}{y}=\int e{^{-x^2}}dx$

But $\int e^{-x^2}dx$  is a non elementary integral. We can express it as a power series

$e^{-x^2}= 1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!} +\frac{x^8}{4!}-..$

We can approximate it as

$e^{-x^2}= 1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}$

So differential equation transforms to

$\int\frac{du}{y}=\int$ $(1 - \frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!})dx$

=>ln|y| = $x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} + C$

By initial condition y(5)=1

0 = $5 - \frac{5^3}{3}+\frac{5^5}{10}-\frac{5^7}{42} + C$

C = 1584.286

So ln|y| = $x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} + 1584.286$

=> y = $e^{(x - \frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42} + 1584.286)}$

This is the explicit solution of the given differential equation

$\mathbf ALTERNATIVELY$

$\frac{dy}{dx} = ye^{-x^2}$

=> $\frac{dy}{y} = e^{-x^2}dx$

=> $\int\frac{du}{y}=\int e{^{-x^2}}dx$

As $\int e{^{-x^2}}dx$ is not an elementary integral, we will express it as error function

=> ln|y| = ($\frac{\sqrtπ}{2}$ ) erf(x) + C₁, C₁ is integration constant

Where erf(x) = $\int_0^x e{^{-t^2}}dt$

So y = $e^{ (\frac{\sqrtπ}{2}) erf(x) + C_1}$

=> y = C$e^{ (\frac{\sqrtπ}{2}) erf(x)}$

By initial condition y (5)=1

1 = C$e^{ (\frac{\sqrtπ}{2}) erf(5) }$

C = $e^{ -(\frac{\sqrtπ}{2}) erf(x) }$

So y = $e^{ (\frac{\sqrtπ}{2}) erf(x) - (\frac{\sqrtπ}{2}) erf(5) }$

=> y = $e^{ (\frac{\sqrtπ}{2}) [erf(x) - erf(5) ]}$

This is the explicit solution of the given differential equation