Solve the Differential equation $left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin...
Solve the equation
$$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$
My try:
Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$
so have modified the equation as:
$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$
$$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$
$$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$
any way to proceed here?
algebra-precalculus differential-equations derivatives indefinite-integrals
asked Nov 29 at 8:26
Umesh shankar
2,55321219
add a comment |
Solve the equation
$$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$
My try:
Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$
so have modified the equation as:
$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$
$$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$
$$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$
any way to proceed here?
algebra-precalculus differential-equations derivatives indefinite-integrals
asked Nov 29 at 8:26
Umesh shankar
2,55321219
add a comment |
Solve the equation
$$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$
My try:
Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$
so have modified the equation as:
$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$
$$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$
$$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$
any way to proceed here?
algebra-precalculus differential-equations derivatives indefinite-integrals
asked Nov 29 at 8:26
Umesh shankar
2,55321219
Solve the equation
$$left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy=0$$
My try:
Letting $e^{sin x}=t$ we have $e^{sin x}cos xdx=dt$
so have modified the equation as:
$$ydt+2tdy-y^3dx-3y^2xdy+2xydx+x^2dy-y^2xdy+x^2dy-4y^2dy=0$$ $implies$
$$ydt+tdy+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$ $implies$
$$d(ty)+tdy-dleft(y^3xright)+dleft(x^2yright)+(x^2-xy^2)dy=4y^2dy$$
any way to proceed here?
algebra-precalculus differential-equations derivatives indefinite-integrals
algebra-precalculus differential-equations derivatives indefinite-integrals
asked Nov 29 at 8:26
Umesh shankar
2,55321219
asked Nov 29 at 8:26
Umesh shankar
2,55321219
asked Nov 29 at 8:26
Umesh shankar
2,55321219
asked Nov 29 at 8:26
Umesh shankar
2,55321219
asked Nov 29 at 8:26
Umesh shankar
2,55321219
2,55321219
add a comment |
add a comment |
1 Answer
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If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$
So the solution is
$$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$
So the solution is
$$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
add a comment |
If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$
So the solution is
$$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
add a comment |
If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$
So the solution is
$$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
If you multiply the expression on the left-hand side by $y$, you get the exact differential,
$$ y times left(left(y e^{sin x}cos x-y^3+2xyright)dx+left(2e^{sin x}-4y^2(x+1)+2x^2right)dy right) =dleft( y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2right).$$
So the solution is
$$ y^2 e^{sin x} - y^4 (x + 1) + x^2 y^2 = {rm constant}.$$
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
answered Nov 29 at 8:36
Kenny Wong
16.8k21135
16.8k21135
add a comment |
add a comment |
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