Question

I.In each of Problems 23 through 30, use the method of reduction of order to find a second solution of the given differential equation.

1.  t2y″ − 4ty′ + 6y = 0, y1(t) = t2

2.  xy″ − y′ + 4x3y = 0, x > 0; y1(x) = sin x2

Answer 1

1.

We seek a solution of the form $y(t) = v(t)y_1(t).$ This gives us

$$y(t)=v(t)t^2$$

$$y'(t)=v't^2+2tv$$

$$y''(t)=v''t^2+4tv'+2v$$

Substitute

$$t^4v''+4t^3v'+2t^2v− 4t^3v'-8t^2v+6t^2v = 0$$

$$t^4v'' = 0$$

$$v''=0$$

Integrate

$$v'=c_1$$

Integrate

$$v=c_1t+c_2$$

Therefore, the general solution is

$$y(t)=(c_1t+c_2)t^2$$

$$y(t)=c_1t^3+c_2t^2$$

The second solution is $y_2(t)=t^3.$

2.

We seek a solution of the form $y(x) = v(x)y_1(x).$ This gives us

$$y(x)=v(x)\sin(x^2)$$

$$y'(x)=v'\sin(x^2)+2x\cos(x^2)v$$

$$y''(x)=v''\sin(x^2)+2x\cos(x^2)v'+2\cos(x^2)v$$

$$-4x^2\sin(x^2)v+2x\cos(x^2)v'$$

Substitute

$$x\sin(x^2)v''+4x^2\cos(x^2)v'+2x\cos(x^2)v$$

$$-4x^3\sin(x^2)v-\sin(x^2)v'-2x\cos(x^2)v$$

$$+4x^3\sin(x^2)v(x)=0$$

$$x\sin(x^2)v''+4x^2\cos(x^2)v'-\sin(x^2)v'=0$$

$$x\sin(x^2)v''=(-4x^2\cos(x^2)+\sin(x^2))v'$$

$$\dfrac{v''}{v'}=-4x\cos(x^2)+\dfrac{\sin(x^2)}{x}$$ $$d(\ln(v'))=(-4x\dfrac{\cos(x^2)}{\sin(x^2)}+\dfrac{1}{x})dx$$

Integrate

$$\int d(\ln(v'))=\int (-4x\dfrac{\cos(x^2)}{\sin(x^2)}+\dfrac{1}{x})dx$$

Find

$$-\int4x\dfrac{\cos(x^2)}{\sin(x^2)}dx$$

$$u=\sin(x^2), du=2x\cos(x^2)dx$$

$$-\int4x\dfrac{\cos(x^2)}{\sin(x^2)}dx=-2\int \dfrac{du}{u}=-2\ln(u)+\ln C_1$$

$$=-2\ln(\sin(x^2))+\ln C_1$$

$$\ln(v')=\ln(\sin^{-2}(x^2))+\ln(x)+\ln C_1$$

$$v'=\dfrac{C_1 x}{\sin^2(x^2)}$$

Integrate

$$v=\int \dfrac{C_1 x}{\sin^2(x^2)}dx$$

Find

$$\int\dfrac{C_1 x}{\sin^2(x^2)}dx$$

$$u=x^2, du=2xdx$$

$$\int\dfrac{C_1 x}{\sin^2(x^2)}dx=\dfrac{C_1}{2}\int\dfrac{1}{\sin^2(u)}du$$

$$=C_2\cot(u)+C_3=C_2\cot(x^2)+C_3$$

$$v(x)=C_2\cot(x^2)+C_3$$

Then

$$y(x)=(C_2\cot(x^2)+C_3)\sin(x^2)$$

$$=C_2\cos(x^2)+C_3\sin(x^2)$$

Therefore, the general solution is

$$y(x)=C_2\cos(x^2)+C_3\sin(x^2)$$

The second solution is $y_2(x)=\cos(x^2).$