Question

suppose the function y=y(x) solves the initial value problem
dy/dx=2y/1+x^2
y(0)=2
find y(2)

Answer 1

Answer:

$y(2)=2e^{2\tan^{-1}(2)}$

Step-by-step explanation:

Given the initial value problem.

$\frac{dy}{dx}=\frac{2y}{1+x^2} ; \ y(0)=2$

Find y(2)

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$\boxed{\left\begin{array}{ccc}\text{\underline{Seperable Differential Equation:}}\\\frac{dy}{dx} =f(x)g(y)\\\\\rightarrow\int\frac{dy}{g(y)}=\int f(x)dx \end{array}\right }$

(1) - Solving the separable DE

$\frac{dy}{dx}=\frac{2y}{1+x^2} \\\\\Longrightarrow \frac{1}{y}dy =\frac{2}{1+x^2}dx\\ \\\Longrightarrow \int \frac{1}{y}dy =2 \int\frac{1}{1+x^2}dx\\\\\Longrightarrow \boxed{ \ln(y)=2\tan^{-1}(x)+C}$

(2) - Find the arbitrary constant "C" with the initial condition

$\text{Recall} \rightarrow y(0)=2\\ \\ \ln(y)=2\tan^{-1}(x)+C\\\\\Longrightarrow \ln(2)=2\tan^{-1}(0)+C\\\\\Longrightarrow \ln(2)=0+C\\\\\therefore \boxed{C=\ln(2)}$

(3) - Form the solution

$\boxed{\boxed{ \ln(y)=2\tan^{-1}(x)+\ln(2)}}$

(4) - Solve for y

$\ln(y)=2\tan^{-1}(x)+\ln(2)\\\\ \Longrightarrow \ln(y)-\ln(2)=2\tan^{-1}(x)\\\\ \Longrightarrow \ln(\frac{y}{2} )=2\tan^{-1}(x)\\\\ \Longrightarrow e^{\ln(\frac{y}{2} )}=e^{2\tan^{-1}(x)}\\\\ \Longrightarrow \frac{y}{2} =e^{2\tan^{-1}(x)}\\\\\therefore \boxed{y=2e^{2\tan^{-1}(x)}}$

(5) - Find y(2)

$y=2e^{2\tan^{-1}(x)}\\\\\therefore \boxed{\boxed{y(2)=2e^{2\tan^{-1}(2)}}}$

Thus, the problem is solved.