Answer to Question #203719 in Differential Equations for irfana Ashraf

Question

y^''-x²y'-2xy=0;y(0)=1&y'(0)=0 solve by reduction of order

Answer 1

$\text{We can re-write the equation as: }\\ \frac{d^{2} y(x)}{d x^{2}}-x^{2} \frac{d y(x)}{d x}-2 x y(x)=0, \text{ such that }y(0)=1 \text{ and } y^{\prime}(0)=0\\ \text{We solve as follows: }\\ \text{Substitute }-2 x=\frac{d}{d x}\left(-x^{2}\right) :\\ \frac{d^{2} y(x)}{d x^{2}}-x^{2} \frac{d y(x)}{d x}+\frac{d}{d x}\left(-x^{2}\right) y(x)=0\\ \text{Apply the reverse product rule }f \frac{d g}{d x}+g \frac{d f}{d x}=\frac{d}{d x}(f g)\text{ to the left-hand side and simplify:}\\ \frac{d}{d x}\left(\frac{d y(x)}{d x}\right)+\frac{d}{d x}\left(-x^{2} y(x)\right)=0\\ \text{Factor: }\\ \frac{d}{d x}\left(\frac{d y(x)}{d x}-x^{2} y(x)\right)=0\\ \text{Integrate both sides with respect to }x :\\ \int \frac{d}{d x}\left(\frac{d y(x)}{d x}-x^{2} y(x)\right) d x=\int 0 d x\\ \text{Evaluate the integrals: }\\ \frac{d y(x)}{d x}-x^{2} y(x)=c_{1}, \qquad \qquad \cdots (i)\\ (\text{ where }c_{1}\text{ is an arbitrary constant.})$

$\text{Let }\mu(x)=e^{\int-x^{2} d x}=e^{-x^{3} / 3}.\\ \text{Multiply both sides by } \mu(x) :\\ e^{-x^{3} / 3} \frac{d y(x)}{d x}-\left(e^{-x^{3} / 3} x^{2}\right) y(x)=c_{1} e^{-x^{3} / 3}\\ \text{Substitute }-e^{-x^{3} / 3} x^{2}=\frac{d}{d x}\left(e^{-x^{3} / 3}\right):\\ e^{-x^{3} / 3} \frac{d y(x)}{d x}+\frac{d}{d x}\left(e^{-x^{3} / 3}\right) y(x)=c_{1} e^{-x^{3} / 3}\\ \text{Apply the reverse product rule }f \frac{d g}{d x}+g \frac{d f}{d x}=\frac{d}{d x}(f g) \text{ to the left-hand side:}\\ \frac{d}{d x}\left(e^{-x^{3} / 3} y(x)\right)=c_{1} e^{-x^{3} / 3}\\ \text{Integrate both sides with respect to }x :\\ \int \frac{d}{d x}\left(e^{-x^{3} / 3} y(x)\right) d x=\int c_{1} e^{-x^{3} / 3} d x\\ \text{Evaluate the integrals:}\\ e^{-x^{3} / 3} y(x)=-\frac{c_{1} x \Gamma\left(\frac{1}{3}, \frac{x^{3}}{3}\right)}{3^{2 / 3} \sqrt[3]{x^{3}}}+c_{2}, \text{ where }c_{2} \text{ is an arbitrary constant.}$

$\text{Divide both sides by }\mu(x)=e^{-x^{3} / 3} :\\ y(x)=e^{x^{3} / 3}\left(-\frac{c_{1} x \Gamma\left(\frac{1}{3}, \frac{x^{3}}{3}\right)}{3^{2 / 3} \sqrt[3]{x^{3}}}+c_{2}\right)\\ \text{Simplify the arbitrary constants:}\\ \text{The general solution to the differential equation is:}\\ y(x)=e^{x^{3} / 3}\left(\frac{c_{1} x \Gamma\left(\frac{1}{3}, \frac{x^{3}}{3}\right)}{\sqrt[3]{x^{3}}}+c_{2}\right) \qquad \cdots (ii)$

Evaluating the differential equation at y(0) = 1 and y'(0) =0.

From (i) above:,

$\frac{d y(x)}{d x}-x^{2} y(x)=c_{1}\\ 0-0(1)=0=c_1$

From (ii):

$y(x)=e^{x^{3} / 3}\left(\frac{c_{1} x \Gamma\left(\frac{1}{3}, \frac{x^{3}}{3}\right)}{\sqrt[3]{x^{3}}}+c_{2}\right)\\ 1=e^0(0+c_2)\\ 1=c_2$

Therefore, the solution to the given equation at the initial conditions given is:

$y(x)=e^{x^{3} / 3}+1$