MCQ
$(x + 2{y^3})\frac{{dy}}{{dx}} - y = 0$ નો ઉકેલ મેળવો.
- A$y(1 - xy) = Ax$
- ✓${y^3} - x = Ay$
- C$x(1 - xy) = Ay$
- D$x(1 + xy) = Ay$ (કે જ્યાં $A$ એ સ્વૈર અચળાંક છે )
==> $\frac{{dx}}{{dy}} = \frac{{x + 2{y^3}}}{y}$ or $\frac{{dx}}{{dy}} - \frac{x}{y} = 2{y^2}$,
which is a linear equation of the form $\frac{{dx}}{{dy}} + Px = Q$
So, integrating factor $(I.F.)$$ = {e^{ - \int_{}^{} {\frac{1}{y}dy} }}$and solution is
$x\frac{1}{y} = \int_{}^{} {\frac{1}{y}2{y^2}dy + A = {y^2} + A} $ ==> $x = {y^3} + Ay$
==> ${y^3} - x = Ay;$where $A$ can be $ - ve$or $ + ve$.
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$\left(1+\cos ^{2} \theta\right) x+\sin ^{2} \theta y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\left(1+\sin ^{2} \theta\right) y+4 \sin 3 \theta z=0$
$\cos ^{2} \theta x+\sin ^{2} \theta y+(1+4 \sin 3 \theta) z=0$
ને શૂન્યતર ઉકેલ ધરાવે છે તો $\theta$ ની કિમંત મેળવો.
${b_1} = b - \frac{{b.a}}{{|a{|^2}}}a,\,{b_2} = b + \frac{{b.a}}{{|a{|^2}}}a$,${c_1} = c - \frac{{c.a}}{{|a{|^2}}}a - \frac{{c.b}}{{|b{|^2}}}b$,
${c_2} = c - \frac{{c.a}}{{|a{|^2}}}a--\frac{{c.{b_1}}}{{|{b_1}{|^2}}}{b_1}$,
${c_3} = c - \frac{{c.a}}{{|a{|^2}}}a--\frac{{c.{b_2}}}{{|{b_2}{|^2}}}{b_2}$,
${c_4} = a - \frac{{c.a}}{{|a{|^2}}}a$
તો આપેલ ગણ પૈકી . . . એ લંબ થાય.