Question

Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular solution requested (Label the Gen. Sol. and Particular Sol.). Write both solutions in EXPLICIT FORM (solved for y). x dy/dx = x^3 + 2y subject to: y(2) = 6

Answer 1

Answer:

General Solution is $y=x^{3}+cx^{2}$ and the particular solution is  $y=x^{3}-\frac{1}{2}x^{2}$

Step-by-step explanation:

$x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}$

This is a linear diffrential equation of type

$\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)$..................(i)

here $p(x)=\frac{-2}{x}$

$q(x)=x^{2}$

The solution of equation i is given by

$y\times e^{\int p(x)dx}=\int e^{\int p(x)dx}\times q(x)dx$

we have $e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}$

Thus the solution becomes

$\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+c$$y=x^{3}+cx^{2$

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have $6=8+4c$

Thus solving for c we get c = -1/2

Thus particular solution becomes

$y=x^{3}-\frac{1}{2}x^{2}$