Решите, пожалуйста,срочно надо!!!

Question 1

Question 2

Вы уверены что все правильно записано? Первые два вызывают сомнение

Question 3

Решите задачу:

$(x+y-2)dx-(3x-7y-3)dy=0|*\frac{1}{(3x-7y-3)dx}\\\frac{dy}{dx}=\frac{x-y-2}{3x-7y-3}\\\mathcal4= \left[\begin{array}{cc}1&-1\\3&-7\end{array}\right]=-7+3=-4\\\begin{cases}a+b-2=0|*3\\3a-7b-3=0\end{cases}=\ \textgreater \ \begin{cases}3a+3b-6=0\\-\\3a-7b-3=0\end{cases}\\10b=3=\ \textgreater \ b=\frac{3}{10}\\a+\frac{3}{10}-2=0\\a=\frac{17}{10}\\x=\hat{x}+a=\hat{x}+\frac{17}{10}\\y=\hat{y}+b=\hat{y}+\frac{3}{10}$
$\frac{d(\hat{y}+\frac{3}{10})}{d(\hat{x}+\frac{17}{10})}=\frac{\hat{x}+\frac{17}{10}+\hat{y}+\frac{3}{10}-2}{3\hat{x}+\frac{51}{10}-7\hat{y}+\frac{21}{10}-3}\\\frac{d\hat{y}}{d\hat{x}}=\frac{\hat{x}+\hat{y}}{3\hat{x}-7\hat{y}}\\\hat{y}=t\hat{x}=\ \textgreater \ \hat{y}'=t'\hat{x}+t\\t'\hat{x}+t=\frac{\hat{x}+t\hat{x}}{3\hat{x}-7t\hat{x}}\\\frac{\hat{x}dt}{d\hat{x}}+t=\frac{1+t}{3-7t}\\\frac{\hat{x}dt}{d\hat{x}}=\frac{1+t}{3-7t}-t\\\frac{\hat{x}}{d\hat{x}}=\frac{7t^2-2t+1}{(3-7t)dt}$
$\frac{d\hat{x}}{\hat{x}}=-\frac{(7t-3)dt}{7t^2-2t+1}\\\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}\frac{(14t-2-4)dt}{7t^2-2t+1}\\\int\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}(\int\frac{(14t-2)dt}{7t^2-2t+1}-4\int\frac{dt}{7t^2-2t+1})\\\int\frac{d\hat{x}}{\hat{x}}=-\frac{1}{2}\int\frac{d(7t^2-2t+1)}{7t^2-2t+1}+\frac{2}{\sqrt7}\int\frac{d(\sqrt7t-\frac{1}{\sqrt7})}{(\sqrt7t-\frac{1}{\sqrt7})^2+\frac{6}{7}}\\ln|x|=-\frac{1}{2}ln|7t^2-2t+1|+\frac{2}{3}arctg\frac{7t-1}{\sqrt6}+C$
$ln|\hat{x}|+\frac{1}{2}ln|7t^2-2t+1|-\frac{2}{3}arctg\frac{7t-1}{\sqrt6}=C\\ln|\hat{x}|+\frac{1}{2}ln|\frac{7\hat{y}^2}{\hat{x}^2}-2\frac{\hat{y}}{\hat{x}}+1|-\frac{2}{3}arctg\frac{\frac{7\hat{y}}{\hat{x}}-1}{\sqrt6}=C\\ln|x-\frac{17}{10}|+\frac{1}{2}ln|\frac{7(y-\frac{3}{10})^2}{(x-\frac{17}{10})^2}-2\frac{y-\frac{3}{10}}{x-\frac{17}{10}}+1|-\frac{2}{3}arctg\frac{\frac{7(y-\frac{3}{10})}{x-\frac{17}{10}}-1}{\sqrt6}=C$
$ln|x-\frac{17}{10}|+\frac{1}{2}ln|\frac{700y^2+100x^2-80y-200xy-280x+250}{100x^2-340x+289}|-\\-\sqrt\frac{2}{3}arctg\frac{70y-10x-4}{\sqrt6(10x-17)}=C$

$(3x^2+6xy^2)dx+(6x^2y+4y^3)dy=0\\\frac{\delta P}{\delta y}=12xy\\\frac{\delta Q}{\delta x}=12xy\\\frac{\delta P}{\delta y}=\frac{\delta Q}{\delta x}\\\frac{\delta F}{\delta x}=(3x^2+6xy^2)\ ;\frac{\delta F}{dy}=(6x^2y+4y^3)\\\frac{\delta F}{\delta x}dx+\frac{\delta F}{\delta y}dy=0\\F=\int(3x^2+6xy^2)dx=3\int x^2dx+6y^2\int xdx=x^3+3y^2x^2+\phi(y)\\\frac{\delta F}{\delta{y}}=(x^3+3y^2x^2+\phi(y))'=6x^2y+\phi'(y)\\6x^2y+\phi'(y)=6x^2y+4y^3\\\phi'(y)=4y^3\\\phi(y)=4\int y^3dy=y^4+C\\F=x^3+3y^2x^2+y^4+C$
$x^3+3y^2x^2+y^4=C$

$y'''-3y''+3y'-y=0\\\lambda^3-3\lambda^2+3\lambda-1=0\\\lambda=1;1-3+3-1=0\\\lambda^3-3\lambda^2+3\lambda-1:(\lambda-1)=\lambda^2-2\lambda+1\\\lambda^3-3\lambda^2+3\lambda-1=(\lambda-1)(\lambda-1)^2=0\\\lambda_{1,2,3}=1\\y=C_1e^x+xC_2e^x+x^2C_3e^x$