Solve the differential equation $t y''-y'+4t^3y=0$

Question

Solve the differential equation $t y''-y'+4t^3y=0$ using method reduction where $y_1=\sin (t^2)$

my attempt:

$y_2=vy_1=v\sin (t^2)$ $y_2'=v'\sin(t^2)+2t\cos(t^2)\\ y_2''=v''\sin (t^2)+4tv\cos(t^2)+2v\cos(t62)4t^2v\sin(t^2)$

Hence $t y_2''-y_2'+4t^3y_2=0\\ tv''\sin (t^2)+4t^2v\cos(t^2)+2vt\cos(t^2)+4t^3v\sin(t^2)-v'\sin(t^2)-2t\cos(t^2)-4t^3v\sin(t^2)=0\\ v''t\sin(t^2)+4t^2v'\cos(t^2)-v'\sin(t^2)=0$

how to proceed from here?

Answer 1

Susbstitute $y=v'$ thats a first order equation

$$v''t\sin(t^2)+4t^2v'\cos(t^2)-v'\sin(t^2)=0$$ $$y't\sin(t^2)+y(4t^2\cos(t^2)-\sin(t^2))=0$$ $$y't\sin(t^2)=-y(4t^2\cos(t^2)-\sin(t^2))$$ It's seperable. $$ \ln|y|=-\int \frac {(4t^2\cos(t^2)-\sin(t^2))}{t\sin(t^2)}dt$$ $$ \ln|y|=-\int \frac {(4t\cos(t^2)}{\sin(t^2)}dt+\ln|t|$$ Substitute $u =\sin(t^2) \implies du=2t\cos(t^2)dt$ $$ \ln|y|=-2\int \frac {du}{u}+\ln|t|$$ $$ \ln|y|=-2\ln|\sin(t^2)|+\ln|t|+K$$ $$ y=(\sin(t^2))^{-2} K t$$ $$ v'=((\sin(t^2))^{-2} K t$$ $$ v=K\int \frac {tdt} {\sin^2(t^2)}$$ Substitute $u=t^2 \implies du=2tdt$ $$ v=K\int \frac {du} {\sin^2(u)}$$ You should get $$\boxed {v=\frac {K_1} {\tan(t^2)}+K_2}$$ And finaly: $$\boxed {y(t)=K_1{\cos(t^2)}+ {K_2} {\sin(t^2)}}$$

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Hint

I used this in the final step... $$ I=\int \frac {du} {\sin^2(u)}=\int \frac {\cos^2(u)du} {\cos^2(u)\sin^2(u)}=\int \frac {d(\tan(u))} {\tan^2(u)}=-\frac 1 {\tan(u)}$$

Answer 2

Let $w = v^\prime.$ What you have is a first order ODE $$w^\prime t \sin(t^2) + (4 t^2 \cos(t^2) - \sin(t^2)) w = 0,$$ so

$$\log w(x) = \int_1^x \frac{(4 t^2 \cos(t^2) - \sin(t^2))}{t \sin(t^2)} d t = \int_1^x 4 t \cot t^2 d t - \int_1^t \frac1t d t.$$

I leave both integrals on the RHS to you.