Rozwiązać równania

[zaliczenie14]

Proszę o sprawdzenie zadań.

Rozwiązać równania:
a) \(\displaystyle{ y'=-2 \sqrt{x} \cdot \sin x }\)
b) \(\displaystyle{ y'+1 y^{2} \cdot \sin x=3(xy)^{2} }\)
\(\displaystyle{ y(0)=1|| y(0)=0}\)

a)
\(\displaystyle{ \frac{dy}{dx}=-2 \sqrt{x} \cdot \sin x }\)

\(\displaystyle{ dy=-2 \sqrt{x} \cdot \sin x dx}\)

\(\displaystyle{ \frac{dy}{ \sqrt{y} }=-2\sin x dx }\)
\(\displaystyle{ \int_{}^{} \frac{dy}{ \sqrt{y} }= \int_{}^{} -2\sin x }\)
\(\displaystyle{ \int_{}^{} y^{-1/2} dy=-2 \int_{}^{} \sin x dx }\)

\(\displaystyle{ 2 \sqrt{y} =2\cos x+C }\)
\(\displaystyle{ y=(\cos x+ \frac{1}{2}C)^{2} }\)


b)
\(\displaystyle{ y'= 3x^{2} y^{2}- y^{2}\sin x }\)
\(\displaystyle{ y'= y^{2}( 3x^{2}-\sin x) }\)
\(\displaystyle{ \frac{dy}{dx}= y^{2}( 3x^{2}-\sin x) }\)

\(\displaystyle{ dy= y^{2}( 3x^{2}-\sin x) dx}\)

\(\displaystyle{ \frac{dy}{ y^{2} }= 3x^{2}-\sin x dx }\)

\(\displaystyle{ \int_{}^{} \frac{dy}{ y^{2} }= \int_{}^{} 3x^{2}-\sin x dx }\)
\(\displaystyle{ \int_{}^{} y^{-2} dy= x^{3}+\cos x +C }\)
\(\displaystyle{ -\frac{1}{y}= x^{3}+\cos x +C }\)
\(\displaystyle{ y= \frac{-1}{ x^{3}+\cos x +C} }\)

dla \(\displaystyle{ y(0)=1}\)
\(\displaystyle{ 1= \frac{-1}{1+C} }\)
\(\displaystyle{ C=-2}\)